C++博客-Programming Every Day-随笔分类-POJ

poj 1273 Drainage Ditches

田兵 — Tue, 31 Aug 2010 14:45:00 GMT

1273:Drainage Ditches

时间限制:: 1000ms
内存限制:: 10000kB

描述

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch.
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network.
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle.

输入

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

输出

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

样例输入

样例输出

最大流问题，注意有重边，要 cap[s][e]+=c;

#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int MAX=205,INF=0x7fffffff/10;
int cap[MAX][MAX];
int flow[MAX][MAX];
int pre[MAX], m[MAX];
int N, M;

int bfs(int s)
{
    memset(m,0,sizeof m);
    memset(pre,0,sizeof pre);
    
    m[1]=INF;
    queue<int> Q;
    Q.push(s);
    while(!Q.empty())
    {
        int u=Q.front(); Q.pop();
        
        for(int i=1; i<=N; i++)
        {
                if(i==u)continue;
                if(m[i]==0&&cap[u][i]>flow[u][i])
                {
                    pre[i]=u;
                    m[i]=( cap[u][i]-flow[u][i]>m[u]?m[u]:cap[u][i]-flow[u][i] );
                    Q.push(i);
                }
        }                
    }
    if(m[N]==0)return 0;
    
    int temp=N,tt;
    while(temp!=1)
    {
             tt=pre[temp];
             flow[tt][temp]+=m[N];
             flow[temp][tt]-=m[N];
             temp=tt;
    }
    return m[N];
}

int main()
{
   
    while(cin>>M>>N)
    {
         memset(cap,0,sizeof cap);
         memset(flow, 0,sizeof flow);
         int s,e,c;
         for(int i=1; i<=M; i++)
         {
                 cin>>s>>e>>c;
                 cap[s][e]+=c;
         }
         
         int ans=0;
         while(1)
         {
              int temp=bfs(1);
              if(temp==0)break;
              ans+=temp;
         }
         cout<<ans<<endl;
    }
    
    system("pause");
    return 0;
}

田兵 2010-08-31 22:45 发表评论

poj 2377 Bad Cowtractors 最大生成树

田兵 — Sun, 29 Aug 2010 03:31:00 GMT

注意重边，还有-1的情况

这题可以增强对最小生成树的理解，同样的方法，只是每次取最大的边。

#include<iostream>
using namespace std;

const int INF=0x7ffffff;
const int MAX=1001;
int mp[1001][1001]={0};
int d[MAX]={0};
bool f[MAX]={0};
int Prim(int N)
{
    int ans=0;
    f[1]=true;
    for(int i=2; i<=N; i++)
            d[i]=(mp[1][i]==0?INF:mp[1][i]);
    
    for(int i=1; i<N; i++)
    {
            int max=-1,u;
            for(int j=1; j<=N; j++)
            {
                    if(f[j]==false&&d[j]!=INF&&d[j]>max)
                    {
                                max=d[j];
                                u=j;
                    }
            }
            
            if(max==-1)
            {
                       return -1;
            }
            ans+=max;
            f[u]=true;
            for(int j=1; j<=N; j++)
                    if(f[j]==false&&mp[u][j]!=0&&(d[j]==INF||mp[u][j]>d[j]))
                    d[j]=mp[u][j];
            
    }
    
    
    return ans;
}

int main()
{
    int N, M;
    cin>>N>>M;
    int a,b,c;
    for(int i=1; i<=M; i++)
    {
            cin>>a>>b>>c;
            if(mp[a][b]<c)
            mp[a][b]=mp[b][a]=c;
    }   
   
    int ans=Prim(N);    
    cout<<ans<<endl;
    system("pause");
    return 0; 
}

田兵 2010-08-29 11:31 发表评论

poj 1338 Ugly Numbers

田兵 — Thu, 26 Aug 2010 02:11:00 GMT

用Dev发生了一件很神奇的事，它竟然输出：
Language:DefaultUgly Numbers
请按任意键继续. . .

该语句我程序里绝对没有，并且还知道是丑数……
本来uglynum[]数组是定义在min前面的，挪到main函数里就没那个输出了。

还有一点诡异的地方：
看下面的程序第一个while里面我调试的cout<

#include<iostream>
using namespace std;

int min(int a, int b){ return a>b?b:a; }
int main()
{
    int uglynum[2000]={1,1};
    int cnt2=0,cnt3=0,cnt5=0;
    for(int i=1; i<=1501; i++)
    {
           while( uglynum[cnt2]*2 <= uglynum[i-1] )
                 { cnt2++;cout<<cnt2<<endl;}
           while( uglynum[cnt3]*3 <= uglynum[i-1] )
                  cnt3++;
           while( uglynum[cnt5]*5 <= uglynum[i-1] )
                  cnt5++;
           int mindigit=min(uglynum[cnt2]*2,min(uglynum[cnt3]*3,uglynum[cnt5]*5));
           
           if(mindigit==uglynum[cnt2]*2){ uglynum[i]=uglynum[cnt2]*2 ;cnt2++;}
           if(mindigit==uglynum[cnt3]*3){ uglynum[i]=uglynum[cnt3]*3 ;cnt3++;}
           if(mindigit==uglynum[cnt5]*5){ uglynum[i]=uglynum[cnt5]*5 ;cnt5++;}
    }
    
    int n;
    while(cin>>n&&n!=0)
    {
        cout<<uglynum[n-1]<<endl;
    }
    
    system("pause");
    return 0;
}

最终的代码：

#include<iostream>
using namespace std;

int min(int a, int b){ return a>b?b:a; }
int main()
{
    int uglynum[2000]={1,1};
    int cnt2=0,cnt3=0,cnt5=0;
    for(int i=1; i<=1501; i++)
    {
           while( uglynum[cnt2]*2 <= uglynum[i-1] )
                 { cnt2++;}
           while( uglynum[cnt3]*3 <= uglynum[i-1] )
                  cnt3++;
           while( uglynum[cnt5]*5 <= uglynum[i-1] )
                  cnt5++;
           int mindigit=min(uglynum[cnt2]*2,min(uglynum[cnt3]*3,uglynum[cnt5]*5));
           uglynum[i]=mindigit;
    }
    
    int n;
    while(cin>>n&&n!=0)
    {
        cout<<uglynum[n-1]<<endl;
    }
    
    system("pause");
    return 0;
}

田兵 2010-08-26 10:11 发表评论

poj 1017 Packets

田兵 — Tue, 24 Aug 2010 03:03:00 GMT

花了两个多小时才搞出来。

而且代码写的也挺烦的，就是从大到小放，先放6，再放5 ……，剩下的部分放其他小的正方形。

主要就是放3的时候，如果剩下一个3*3可以放1个2*2
剩下两个3*3可以放3个2*2
剩下3个3*3可以放5个2*2

#include<iostream>
#include<fstream>
using namespace std;

int num[7];
int main()
{
    //ifstream cin("data.txt");
    //ofstream cout("out.txt");  //40 942 264 8 6 5
    for(;;)
    {
        int i;
        for(i=1; i<=6; i++)
            cin>>num[i];
        
        if(num[1]==0&&num[2]==0&&num[3]==0&&num[4]==0&&num[5]==0&&num[6]==0)
            break;
        
        int ans=num[6];//6
        ans+=num[5];//5
        if(num[1]>=num[5]*11)  num[1]-=num[5]*11;
        else num[1]=0;

        if(num[4])
        {
            ans+=num[4];//4
            if(num[2]>=num[4]*5)num[2]-=num[4]*5;
            else  
            {
                int temp=4*(num[4]*5-num[2]); //剩下的装1
                num[2]=0;
                if(num[1]>=temp)  num[1]-=temp;
                else num[1]=0;
            } 
        }
        
        //3
         if(num[3])
         {
            ans+=(num[3]+3)/4;
            int box3=4-num[3]%4; //剩下的3*3的个数
            if(box3==4)box3=0;
            if(box3==2)
            {
                int t;
                if(num[2]>3){ num[2]-=3; t=3;}
                else {  t=num[2]; num[2]=0;}
                if(num[1]>=box3*9-t*4)num[1]-=( box3*9-t*4) ;
                else num[1]=0;
            }
            else if(box3!=0) 
            {
                if(box3==1)
                {
                    if(num[2]==0)
                    {
                        if(num[1]>=9)num[1]-=9;
                        else num[1]=0;
                    }
                    else
                    {
                        num[2]-=1;
                        if(num[1]>=5)num[1]-=5;
                        else num[1]=0;
                    }
                }
                else  //box3==3
                {
                    int tt;
                    if(num[2]>=5){  num[2]-=5; tt=5;}
                    else { tt=num[2]; num[2]=0; } 
                    if(num[1]>=box3*9-tt*4)num[1]-=(box3*9-tt*4);
                    else num[1]=0;
                }
            }
            
         }
    
         //2
         if(num[2])
         {
          ans+=(num[2]+8)/9;
          int box2=9-num[2]%9;
          if(box2==9)box2=0;
          if(num[1]>=box2*4)num[1]-=box2*4;
          else num[1]=0;
         }
         //1
         if(num[1])
         {
                ans+=(num[1]+35)/36;
         }
         
         cout<<ans<<endl;
    }

    return 0;
}

田兵 2010-08-24 11:03 发表评论

poj 2553 The Bottom of a Graph

田兵 — Mon, 23 Aug 2010 12:21:00 GMT

Accepted

1004K

297MS

C++

1585B

求出度为0的强连通分量，可能有多个，按节点序号从小到大输出就行了。

还是用两次DFS 加名字不好记加还不知道怎么发音的的那个算法，第一次原图DFS确定访问顺序，第二次对翻转过边的方向得到的新图DFS

确定强连通分量。

还不太熟，写了一些时间。

#include<iostream>
#include<vector>
using namespace std;
vector<int> adj[5001];
vector<int> adj_op[5001];
bool visit[5001];
int father[5001],outagree[5001];
vector<int> finish;
int n,m;
void dfs1(int v)
{
    visit[v]=true;
     for(int i=0; i<adj[v].size(); i++)
         if(!visit[adj[v][i]])
             dfs1(adj[v][i]);
    finish.push_back(v);
}

void dfs2(int v, int fa)
{
    visit[v]=true;
    for(int i=0; i<adj_op[v].size(); i++)
        if(!visit[adj_op[v][i]])
        {
             father[adj_op[v][i]]=fa;
             dfs2(adj_op[v][i],fa);
        }

}
int main()
{
    while(cin>>n)
    {
        if(n==0)break;
        cin>>m;
        int i,j,s,t;
        for(i=1; i<=n; i++)
        {
            adj[i].clear();
            adj_op[i].clear();
        }

        for(i=1; i<=m; i++)
        {
            cin>>s>>t;
            adj[s].push_back(t);
            adj_op[t].push_back(s);
        }
        
        finish.clear();
        memset(visit, 0, sizeof visit);
        for(i=1; i<=n; i++)
            if(!visit[i])
                dfs1(i);

        memset(visit, 0, sizeof visit);
        memset(father,0,sizeof father);
        int cnt=0;
        for(i=finish.size()-1; i>=0; i--)
            if(!visit[finish[i]])
            {
                cnt++;
                father[finish[i]]=cnt;
                dfs2(finish[i],cnt);
            }
          
         //for(i=1; i<=n; i++)
        //     cout<
         memset(outagree,0,sizeof outagree);
        
         for(i=1; i<=n; i++)
             for(j=0; j<adj[i].size(); j++)
             {
                 if(father[i]!=father[ adj[i][j] ])
                     outagree[ father[i] ]++;
             }
        
             int tt=0;
             for(i=1; i<=n; i++)
                 if(outagree[father[i]]==0)
                 {
                     if(tt!=0)cout<<' ';
                     cout<<i;
                     tt++;
                 }
             cout<<endl;
    }

    return 0;
}

田兵 2010-08-23 20:21 发表评论

poj 2453 An Easy Problem

田兵 — Mon, 23 Aug 2010 10:38:00 GMT

牛人的代码：

#include <stdio.h>
int main()
{
    int n,x;
    while(scanf("%d",&n),n)
    {
        x=n&-n;
        printf("%d\n",n+x+(n^n+x)/x/4);
    }
}

自己写的非常烦的位运算：
一定要注意优先级问题，位运算操作的优先级貌似都很低。

#include<stdio.h>
#include<cmath>
int main()
{
    int I;
//    printf("%d\n",3&4);
    while(scanf("%d",&I)!=EOF&&I!=0)
    {
    
        int ans=(I&(-I));
        if(I-ans==0)
        {
            printf("%d\n",(I<<1));
            continue;
        }

        int t=0,i;
        for( i=ans; ; i=i<<1,t++)
        {
        //    printf("i  %d   %d\n",i,I&i);
            if( (I&i) == 0)break;
        }

        int temp=0;
        for(int j=1; j<=t-1; j++)
        {
            temp=((temp<<1)+1);
        }
        int k=0;
        for(k=1; ; k++)
        {
            if((1<<k)==i)break;
        }
        printf("%d\n",((I>>k)<<k)+i+temp);
    }

    return 0;
}

田兵 2010-08-23 18:38 发表评论

poj 3170 Knights of Ni

田兵 — Sun, 22 Aug 2010 07:54:00 GMT

Knights of Ni

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 1043		Accepted: 440

Description

Bessie is in Camelot and has encountered a sticky situation: she needs to pass through the forest that is guarded by the Knights of Ni. In order to pass through safely, the Knights have demanded that she bring them a single shrubbery. Time is of the essence, and Bessie must find and bring them a shrubbery as quickly as possible.

Bessie has a map of of the forest, which is partitioned into a square grid arrayed in the usual manner, with axes parallel to the X and Y axes. The map is W x H units in size (1 <= W <= 1000; 1 <= H <= 1000).

The map shows where Bessie starts her quest, the single square where the Knights of Ni are, and the locations of all the shrubberies of the land. It also shows which areas of the map can be traverse (some grid blocks are impassable because of swamps, cliffs, and killer rabbits). Bessie can not pass through the Knights of Ni square without a shrubbery.

In order to make sure that she follows the map correctly, Bessie can only move in four directions: North, East, South, or West (i.e., NOT diagonally). She requires one day to complete a traversal from one grid block to a neighboring grid block.

It is guaranteed that Bessie will be able to obtain a shrubbery and then deliver it to the Knights of Ni. Determine the quickest way for her to do so.

Input

Line 1: Two space-separated integers: W and H.

Lines 2..?: These lines describe the map, row by row. The first line describes the most northwest part of the map; the last line describes the most southeast part of the map. Successive integers in the input describe columns of the map from west to east. Each new row of a map's description starts on a new input line, and each input line contains no more than 40 space-separated integers. If W <= 40, then each input line describes a complete row of the map. If W > 40, then more than one line is used to describe a single row, 40 integers on each line except potentially the last one. No input line ever describes elements of more than one row.

The integers that describe the map come from this set:
0: Square through which Bessie can travel
1: Impassable square that Bessie cannot traverse
2: Bessie's starting location
3: Location of the Knights of Ni
4: Location of a shrubbery

Output

Line 1: D, the minimum number of days it will take Bessie to reach a shrubbery and bring it to the Knights of Ni.

Sample Input

8 4
4 1 0 0 0 0 1 0
0 0 0 1 0 1 0 0
0 2 1 1 3 0 4 0
0 0 0 4 1 1 1 0

Sample Output

Hint

Explanation of the sample:

Width=8, height=4. Bessie starts on the third row, only a few squares away from the Knights.

Bessie can move in this pattern to get a shrubbery for the Knights: N, W, N, S, E, E, N, E, E, S, S. She gets the shrubbery in the northwest corner and then makes her away around the barriers to the east and then south to the Knights.

Source

USACO 2005 December Silver

题意： Bessie要到shrubbery的位置，然后把shrubbery带到Ni，Ni、Bessie只有一个，而shrubbery有多个。

那么分别从Bessie 和 Ni的位置用BFS求两次最短路径，然后去二者到shrubbery(也就是map[i][j]==4的(i,j))的位置

的最短路径和的最小值就是答案。

要用scanf 用cin的话八成会超时。

#include<iostream>
#include<stdio.h>
#include<cstring>
#include<queue>
using namespace std;

int W,H,Bessiex,Bessiey,Nix,Niy;
int map[1005][1005],dBessie[1005][1005],d[1005][1005];
bool visit[1005][1005];
int dx[5]={0,0,0,1,-1},
    dy[5]={0,1,-1,0,0};

struct coordinate
{
         int x,y;
};

void BFS(int i, int j)
{
     memset(d, 0, sizeof d);
     memset(visit, 0, sizeof visit);
     queue<coordinate> Q;  
     coordinate temp;
     temp.x=i; temp.y=j;
     Q.push(temp);
     visit[i][j]=1;
     d[i][j]=0;
     while(Q.size())
     {
         temp=Q.front();
         Q.pop();
         int x=temp.x, y=temp.y;
         for(int i=1; i<=4; i++)
         {
                int tx=x+dx[i], ty=y+dy[i];
                if(tx>=1&&tx<=W&&ty>=1&&ty<=H&&!visit[tx][ty]&&map[tx][ty]!=1)
                {
                     d[tx][ty]=d[x][y]+1;
                     visit[tx][ty]=1;
                     coordinate tt;  tt.x=tx; tt.y=ty;
                     Q.push(tt);
                }
         }
                       
     }
}

int main()
{
    scanf("%d %d",&H,&W);//cin>>H>>W;
    memset(dBessie,0,sizeof dBessie);

    for(int i=1; i<=W; i++)
            for(int j=1; j<=H; j++)
            {
                scanf("%d",&map[i][j]);//  cin>>map[i][j];
                  if(map[i][j]==2)
                  {
                                  Bessiex=i;
                                  Bessiey=j;
                  }
                  if(map[i][j]==3)
                  {
                                  Nix=i;
                                  Niy=j;
                  }           
            }
    
    BFS(Bessiex,Bessiey);     

    for(int i=1; i<=W; i++)
    for(int j=1; j<=H; j++)
                 dBessie[i][j]=d[i][j];
               
    int ans=0x7fffffff;
        
    BFS(Nix, Niy);
    
    for(int i=1; i<=W; i++)
    for(int j=1; j<=H; j++)
    {
            if(map[i][j]==4&&dBessie[i][j]+d[i][j]<ans&&dBessie[i][j]&&d[i][j])
//这里原忘了判断最短路是否存在wa了一次（&&dBessie[i][j]&&d[i][j])）
                ans=dBessie[i][j]+d[i][j];
    }  
     
   printf("%d\n",ans);// cout<
     
    system("pause");
    return 0;
}

田兵 2010-08-22 15:54 发表评论

poj 2503 Babelfish

田兵 — Sun, 22 Aug 2010 02:16:00 GMT

Accepted

9620K

1204MS

C++

567B

水题，用stl的map 。

#include<iostream>
#include<map>
#include<string>
using namespace std;
int main()
{
     map<string ,string> mp;
     string str1,str2;
     while(cin>>str1)
     {
        if(getchar()=='\n')break;
        cin>>str2;
        mp.insert(make_pair(str2,str1));                       
     }
     map<string,string>::iterator it;
     
     do
     {
              it=mp.find(str1);
              if(it!=mp.end())
                  cout<<it->second<<endl;
              else cout<<"eh"<<endl;
     }while(cin>>str1);
    
    system("pause");
    return 0;
}

田兵 2010-08-22 10:16 发表评论

poj 2457 Part Acquisition

田兵 — Sat, 21 Aug 2010 07:06:00 GMT

有向无权图单源最短问题，用BFS求得。

打印路径，记录父节点即可。

#include<iostream>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
vector<int>adj[1005];
int d[1005];
bool f[1005];
int father[1005];
void BFS(int n)
{
     memset(f,0,sizeof f);
     memset(d,0,sizeof d);
     queue<int>Q;
     d[1]=1;
     d[n]=-1;
     father[1]=0;
     Q.push(1);
     f[1]=true;
     int temp,i;
     while(!Q.empty())
     {
         temp=Q.front();  Q.pop();
         for(i=0; i<adj[temp].size(); i++)
                  if(!f[adj[temp][i] ])
                  {
                      father[adj[temp][i]]=temp;
                      d[adj[temp][i]]=d[temp]+1;
                      f[adj[temp][i]]=true;
                      Q.push(adj[temp][i]);
                  }
                         
     }
     
}
void print(int i)
{
     if(i==0)return ;
        print(father[i]);
     cout<<i<<endl;
}
int main()
{
    int n,m;
    cin>>m>>n;
    for(int i=1,s,t; i<=m; i++)
    {
            cin>>s>>t;
            adj[s].push_back(t);
    }
    
    BFS(n);
    
    cout<<d[n]<<endl;
    if(d[n]!=-1)
    print(n);
                     
    system("pause");
    return 0;
}

田兵 2010-08-21 15:06 发表评论

poj 1256 Anagram [next_permutation的应用]

田兵 — Sat, 21 Aug 2010 01:49:00 GMT

next_permutation是按ascII码排序的，也就是值的大小排序。所以要加个比较函数。

大小写字母只有第6位不一样，当|32后，大写字母将+32变成小写字母的值。

#include<iostream>
#include<string>
#include<cstring>
#include<algorithm>
using namespace std;
bool cmp(const char a, const char b)
{
    if( (a|32)==(b|32) ) return a<b;
    else return (a|32)<( b|32);
}
int main()
{
    int t;
    cin>>t;
    string str;
    while(t--)
    {
              cin>>str;
              sort(str.begin(),str.end(),cmp);
              cout<<str<<endl;
              while( next_permutation(str.begin(), str.end(),cmp) )
                                              cout<<str<<endl;
    }
    system("pause");
    return 0;
}

田兵 2010-08-21 09:49 发表评论

poj 2230 Watchcow

田兵 — Fri, 20 Aug 2010 11:00:00 GMT

Accepted

3380K

1469MS

C++

有向图的欧拉回路问题。

欧拉问题要加个小结。

#include<iostream>
#include<vector>
using namespace std;
struct type
{
   bool visit;
   int dest;     
};
int N,M;
vector<type>E[10005];

void dfs(int i)
{
     
     for(int k=0; k<E[i].size(); k++)
     {
                     if(E[i][k].visit==false)
                     {
                          E[i][k].visit=true;
                            dfs(E[i][k].dest);
                         
                     }
     }cout<<i<<endl;
}

int main()
{
    cin>>N>>M;
    type temp;
    for(int s,t,i=1; i<=M; i++)
    {
            cin>>s>>t;
            temp.visit=false;
            temp.dest=t;
            E[s].push_back(temp);
            temp.dest=s;
            E[t].push_back(temp);
    }
    
    dfs(1);
    
    system("pause");
    return 0;
}

田兵 2010-08-20 19:00 发表评论

poj 1077 Eight 【经典八数码问题】

田兵 — Fri, 20 Aug 2010 03:04:00 GMT

Accepted

3164K

219MS

G++

3644B

经典题感觉就是不一样。

总是花了很长时间浪费在一些无关紧要的地方，把方向理解错了，正好相反的，调了N久。

后来输出的时候，都往反方向输出A了。

用康托展开做哈希函数，直接BFS就可以了

#include
#include
using namespace std;
int fac[10]={1,1,2,6,24,120,720,5040,40320,362880};
void swap(int &a, int &b)
{
     int t=a; a=b; b=t; 
}
int cantor(int x)
{
    int a[10]={0};
    for(int i=9; i>=1; i--)
    {
            a[i]=x%10;
            x/=10;
    }
    int ans=0;
    for(int i=1; i<=9; i++)
    {
            int c=0;
            for(int j=i+1; j<=9; j++)
                    if(a[j]            ans+=c*fac[a[i]-1];       
    }
    return ans;
}
struct type
{
       int last;
       int now;
       int move;
}Q[362885];
bool used[362885]={0};
bool change(int &x, int ch)//方向理解反了，调r时其实是调l
{
     int status[4][4],temp=x;
     int row,column;
     for(int i=3; i>=1; i--)
             for(int j=3; j>=1; j--)
             {
                     status[i][j]=temp%10;
                     temp/=10;
                     if(status[i][j]==9)
                     {  row=i; column=j; }             
             }
     switch(ch)
     {
               case 1: if(column==1)return false;  //r
                         swap(status[row][column], status[row][column-1]); break;
               case 2: if(column==3)return false;  //l
                         swap(status[row][column], status[row][column+1]);break;
               case 3: if(row==3)return false;     //u
                         swap(status[row][column], status[row+1][column]); break;
               case 4: if(row==1) return false;    //d
                         swap(status[row][column], status[row-1][column]); break;
     }
     temp=0;
     for(int i=1; i<=3; i++)
     for(int j=1; j<=3; j++)
             temp=temp*10+status[i][j];
     x=temp;
     return true;
}
int main()
{
    int now=0;
    int goal=0;
    char t;
    for(int i=1; i<=9; i++)
    {
            cin>>t;
            if(t=='x')t='9';
            now=now*10+int(t-'0');
    }        
    int l=1,r=1;
    Q[1].last=0; Q[1].now=now; 
    int x=cantor(now);
    used[x]=1;
    bool find=false;
    while(l<=r&&find==false&&r<=362881)
    {
          int num=Q[l].now, can;
          int temp;
          //system("pause");
          for(int i=1; i<=4; i++)
          {
                  temp=num;
                  if(change(temp,i))
                  {
                                   can=cantor(temp);
                                   if(!used[can])
                                   {
                                                 used[can]=true;      
                                                 r++;
                                                 Q[r].now=temp;
                                                 Q[r].last=l;
                                                 Q[r].move=i;
                                   }
                                   if(can==goal){find=true; break; }
                  }
          } 
         l++;
    }          
    
    if(find==false)cout<<"unsolvable"<    else 
    {
         int ans[1000];  int len=0;
         
         for( ;r!=1; )
         {
              
              len++;
              ans[len]=Q[r].move;
              r=Q[r].last;
         }
         for(int i=len; i>=1; i--)
         {
                 switch(ans[i])
                 {
                                 case 2:cout<<'r';break;//调一下输出顺序A了
                                 case 1:cout<<'l';break;
                                 case 4:cout<<'u';break;
                                 case 3:cout<<'d';break;
                 }
         }
     cout<    }
    
    system("pause");
    return 0;
}

田兵 2010-08-20 11:04 发表评论

poj 2243 Knight Moves

田兵 — Thu, 19 Aug 2010 13:30:00 GMT

63ms

跟USACO上的一样。
d[i][j][][]表示从(i,j)这个点到所有点的最短路径。对于每个(i,j)调用一次BFS求最短路径。

#include<iostream>
#include<queue>
using namespace std;
int d[9][9][9][9];
bool visit[9][9];
struct type
{
    int x, y;     
};
int dx[9]={0,-2, 2,-1, 1,-2,2,-1,1},
    dy[9]={0,-1,-1,-2,-2, 1,1, 2,2};
void BFS(int i, int j)
{
     memset(visit,0,sizeof visit);
     queue<type> Q;
     type temp; 
     temp.x=i; temp.y=j;
     Q.push(temp);
     visit[i][j]=1;
     while(Q.size())
     {
                   type t=Q.front(); Q.pop();
                   int x=t.x, y=t.y;
                   for(int k=1; k<=8; k++)
                   {
                           int tx=x+dx[k], ty=y+dy[k];
                           if(tx>=1&&tx<=8&&ty>=1&&ty<=8&&!visit[tx][ty])
                           {
                               d[i][j][tx][ty]=d[i][j][x][y]+1;
                               visit[tx][ty]=1;
                               temp.x=tx; temp.y=ty;
                               Q.push(temp);
                           }
                   } 
     }
}

int main()
{
    memset(d,0,sizeof d);
    for(int i=1; i<=8; i++)
    for(int j=1; j<=8; j++)
            BFS(i,j);
    //for(int i=1; i<=8; i++,cout<   // for(int j=1; j<=8; j++)
      //      cout<
    char knight_column1,knight_column2;
    int knight_row1, knight_row2;
    while(cin>>knight_column1)
    {
           cin>>knight_row1>>knight_column2>>knight_row2;
           cout<<"To get from "<<knight_column1<<knight_row1<<" to "<<knight_column2
                <<knight_row2<<" takes "<<
                d[knight_row1][int(knight_column1-'a'+1)][knight_row2][int(knight_column2-'a'+1)]
                <<" knight moves."<<endl;     
           
    }
    
    return 0;
}

田兵 2010-08-19 21:30 发表评论

poj 2299 Ultra-QuickSort

田兵 — Thu, 19 Aug 2010 12:27:00 GMT

又见树状数组，又不是自己写的。

参考代码：http://www.cppblog.com/jie414341055/articles/114256.html

参考树状数组介绍：http://kmplayer.javaeye.com/blog/562119

#include<iostream>
#include<algorithm>
using namespace std;
const int MAX=500005;
int c[MAX]={0},n,arr[MAX]={0};
struct digit
{
       int num,id;
}a[MAX];

int lowbit(int t)
{
    return t&(t^(t-1));
}

int sum(int t)
{
    int total=0;
    while(t>=1)
    {
               total+=c[t];
               t-=lowbit(t);
    }
    return total;
}

void inc(int t)
{
     while(t<=n)
     {
                c[t]++;
                t+=lowbit(t);
     }
}

bool cmp(const digit &a ,const digit &b)
{
     return a.num<b.num;
}

int main()
{
    while(cin>>n&&n!=0)
    {
         memset(c,0,sizeof c);
         memset(arr,0,sizeof arr);
         for(int i=1; i<=n; i++)
         {
             cin>>a[i].num; 
             a[i].id=i;
         }
                    
         sort(a+1, a+1+n, cmp);
         
         arr[a[1].id]=1;
         for(int i=2; i<=n; i++)
                 if(a[a[i].id].num!=a[a[i-1].id].num)
                            arr[a[i].id]=i;
                 else arr[a[i].id]=arr[a[i-1].id];
                 
         long long ans=0;
         for(int i=1; i<=n; i++)
                 {
                      inc(arr[i]);
                      ans+=( sum(n)-sum(arr[i]) );      
                 }
         cout<<ans<<endl;
                       
    }
    
    return 0;
}

田兵 2010-08-19 20:27 发表评论

poj 1321 棋盘问题

田兵 — Thu, 19 Aug 2010 02:31:00 GMT

我把所有的棋盘区域记下来，这一只要搜索这些区域即可。

不过这样貌似还没有直接按行搜索快。失败

#include<iostream>
using namespace std;
char map[10][10];
int f[10][10],n,k,ans,num_chess;
struct type
{
       int x,y;
}chess[65];
bool check(int i, int j)//check if it is possible to put it at (i,j)
{
    for(int k=1; k<=n; k++)  
            if(f[i][k]&&k!=j)return false;
    for(int k=1; k<=n; k++)
            if(f[k][j]&&k!=i)return false;
    return true;
}

void dfs(int cnt, int numchess )
{ 
     if(cnt==k){ ans++; return ; }
     if(numchess>num_chess)return ;
     int x=chess[numchess].x,y=chess[numchess].y;
     
     if(check(x,y))
     {
      f[x][y]=1;
      dfs(cnt+1,numchess+1);
     } 
     
     f[x][y]=0;
     dfs(cnt, numchess+1);
}

int main()
{
    while(cin>>n>>k)
    {
         if(n==-1 && k==-1)break;
         memset(map,0,sizeof map);
         memset(f,0,sizeof f);
         ans=0;
         num_chess=0;
         for(int i=1; i<=n; i++)
         for(int j=1; j<=n; j++)
               {
                         cin>>map[i][j];
                         if(map[i][j]=='#')
                         {
                                           num_chess++;
                                           chess[num_chess].x=i; chess[num_chess].y=j; 
                         }
               }
         dfs(0,1);
         cout<<ans<<endl;
    }
    
    system("pause");
    return 0;
}

田兵 2010-08-19 10:31 发表评论

poj 2676 Sudoku

田兵 — Wed, 18 Aug 2010 11:09:00 GMT

终于A了
140ms+

说一下我的思路：

接受数据后，把每个为0的位置记录下来(tatus[])，并计算该位置可能的状态数，然后排序，从状态数最小的开始搜索。也就是仅搜索为0的位置。

不过对于这组数据：
9
000000000
000000000
000000000
000000000
000000000
000000000
000000000
000000000
000000000
我估计我程序半年都出不了结果

check(x,y,p)函数检查(x,y)位置是否可以放p这个值。
cal()函数计算初始可能的状态数。

#include
#include
#include
using namespace std;
int sudo[10][10];
struct type
{
       int x,y,count; 
}status[82];  
int num_status=0;
bool check(int i, int j, int t)//judge if it is possible to put digit t at (i,j)
{
     for(int k=1; k<=9; k++)
             if(sudo[i][k]==t)return false;
     for(int k=1; k<=9; k++)
             if(sudo[k][j]==t)return false;
     int x=(i-1)/3*3+1 ,y=(j-1)/3*3+1;
     for(int s=0; s<=2; s++)
     for(int tt=0; tt<=2; tt++)
             if(sudo[x+s][y+tt]==t)return false;
     return true;             
}
int cal(int i, int j)//calculate the the number of  possible status of (i,j) 
{
    int cnt=0;
    for(int p=1; p<=9; p++)
            if(check(i,j,p))cnt++;
            
    return cnt;
}

bool cmp(const type &a, const type &b)
{
     return a.count}
void dfs(int cnt,bool &find)
{
     if(cnt>num_status)return ;
     if(find){ return ;}
     int x=status[cnt].x; 
     int y=status[cnt].y;
     
     if(cnt==num_status)
     {
              for(int p=1; p<=9; p++)
                      if(check(x,y,p))
                      {
                                      sudo[x][y]=p;
                                      for(int i=1; i<=9; i++)
                                      {
                                              for(int j=1; j<=9; j++)
                                                      cout<                                              cout<                                      }
                                      find=true;
                                      return ;
                      }
                      return ;    
     }
     for(int p=1; p<=9; p++)
     {
             if(check(x,y,p))
             {
              sudo[x][y]=p;
              dfs(cnt+1,find);
             }
            sudo[x][y]=0; //刚开始一直错就因为没置0，check函数当当前位置是0的时候才有效。
     }
         
}
 
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
       memset(sudo,0,sizeof sudo); 
       char ch;
       num_status=0;
       for(int i=1; i<=9; i++)
               for(int j=1; j<=9; j++)
               { 
                cin>>ch;
                sudo[i][j]=int(ch-'0');
                }
                
       for(int i=1; i<=9; i++)
       for(int j=1; j<=9; j++)
                  if(sudo[i][j]==0)
                  {
                      num_status++; 
                      status[num_status].x=i; status[num_status].y=j; 
                      status[num_status].count=cal(i,j); 
                  }
       
       sort(status+1,status+1+num_status,cmp);
      // for(int i=1; i<=num_status ;i++)
        //      cout<<' '<
       
       bool find=false;
       dfs(1,find);
       if(num_status==0)
       {
                    for(int i=1; i<=9; i++)
                    {
                              for(int j=1; j<=9; j++)
                                   cout<                              cout<                    }    
       }
    }
    system("pause");
    return 0;
}

田兵 2010-08-18 19:09 发表评论

poj 2240 Arbitrage 【Floyd】

田兵 — Wed, 18 Aug 2010 07:15:00 GMT

floyd 松弛一下即可。

#include<iostream>
#include<string>
#include<string.h>
using namespace std;
string arr[35];
double adj[35][35];
const double INF=1e30;
void Floyd(int n)
{
     for(int k=1; k<=n; k++)
     for(int i=1; i<=n; i++)
     for(int j=1; j<=n; j++)
     {
             if(adj[i][j]<adj[i][k]*adj[k][j])
                    adj[i][j]=adj[i][k]*adj[k][j];       
     }
}

int main()
{
    int n,m;
    for(int tt=1; ; tt++)
    {
       cin>>n;
       if(n==0)break;
       memset(adj,0,sizeof adj);
                       
       for(int i=1; i<=n; i++)
               cin>>arr[i];
      
       string str1,str2; 
       double rate;
       cin>>m;
       for(int i=1,s,t; i<=m; i++)
       {
               cin>>str1>>rate>>str2;
              
               for(int j=1; j<=n; j++)
                       if(str1==arr[j])
                       { s=j; break; }
               for(int j=1; j<=n; j++)
                       if(str2==arr[j])
                       { t=j; break; }
               adj[s][t]=rate;
       }
       
       Floyd(n);
       
       bool flag=false;
       for(int i=1; i<=n; i++)
               if(adj[i][i]-1>0.00005)flag=true;
       cout<<"Case "<<tt<<": ";
       if(flag)cout<<"Yes"<<endl;
       else cout<<"No"<<endl;
    }
    return 0;
}

田兵 2010-08-18 15:15 发表评论

poj 2421 Constructing Roads prim

田兵 — Wed, 18 Aug 2010 03:38:00 GMT

最小生成树，把已经修建的道路的距离置0，然后求最小生成树，prim用的还不熟

prim好像打错了

#include
using namespace std;
int adj[101][101]={0};
const int INF=0x7fffffff/1000;
int prime(int n)
{
    bool in[101];
    memset(in,0,sizeof in);
    int d[101]={0};
    for(int i=1; i<=n; i++)
            d[i]=INF;
    d[1]=0; 
    int ans=0;
    for(int i=1; i<=n; i++)
    {
            int min=INF,u=0;
            for(int j=1; j<=n; j++)
            {
                    if(!in[j]&&d[j]                    {
                           min=d[j];
                           u=j;             
                    }
            }
            
            ans+=min; 
            in[u]=true;
            
            for(int j=1; j<=n; j++)
                    if(!in[j]&&d[j]>adj[u][j])
                            d[j]=adj[u][j];
    }
    
    return ans;
}
int main()
{
    int n;
    cin>>n;
    for(int i=1; i<=n; i++)
            for(int j=1; j<=n; j++) 
                    cin>>adj[i][j];
    
    int Q,s,t;
    cin>>Q;
    for(int i=1; i<=Q; i++)
           {  cin>>s>>t; adj[s][t]=0; adj[t][s]=0; }
            
    int ans=prime(n);
    cout<    system("pause");
    return 0;
}

田兵 2010-08-18 11:38 发表评论

poj 2186 Popular Cows

田兵 — Tue, 17 Aug 2010 12:07:00 GMT

用adj作为邻接表，存正向的边，把边反向存在adj_op里。

第一次dfs1确定一颗深度优先搜索树，用finish记录访问顺序，然后从finish后面往前进行dfs2，每一次dfs2可以确定一个强连通分量。

证明参考算法导论

#include
#include
using namespace std;
const int MAX=10005;
vectoradj[MAX], adj_op[MAX];
bool visit[MAX]={0},out[MAX];
int sblg[MAX],n,m;
vector finish;
void dfs1(int i)
{
         if(visit[i])return ;
         visit[i]=true;
         for(int k=0; k                 if(!visit[adj[i][k]])dfs1(adj[i][k]);
         finish.push_back(i);                 
}

void dfs2(int i, int c)
{
     if(visit[i])return ;
     visit[i]=true;  
     sblg[i]=c;
     for(int k=0; k             if(!visit[adj_op[i][k]])dfs2(adj_op[i][k],c);
}

int main()
{
    while(cin>>n>>m)
    {
        memset(visit,0,sizeof visit); memset(out,0,sizeof out); memset(sblg,0,sizeof sblg);
        for(int i=1; i<=n; i++){ adj[i].clear(); adj_op[i].clear(); }
        int u,v;
        for(int i=1; i<=m; i++)
               {
                     cin>>u>>v;
                     adj[u].push_back(v);
                     adj_op[v].push_back(u);
               }    
        for(int i=1; i<=n; i++)
                if(!visit[i])dfs1(i);
        memset(visit,0,sizeof visit);
        int cnt=0;
        
        
        for(int i=finish.size()-1; i>=0; i--)
        {
                if(!visit[finish[i]])
                {
                    cnt++;
                    dfs2(finish[i],cnt);
                }
        }
       
        for(int i=1; i<=n; i++)
        {
                for(int j=0; j                {
                        if(sblg[i]!=sblg[adj[i][j]])out[sblg[i]]=true;
                }
        }
        int count=0,index=0,num=0;
        for(int i=1; i<=cnt; i++)
                if(out[i]==0){ count++;index=i; }
        //for(int i=1; i<=n; i++)                
          //      cout<        if(count==1)
        {
                   for(int i=1; i<=n; i++)
                           if(sblg[i]==index)num++;
                   cout<        }
        else cout<<0<            
    }
    
    return 0;
}

田兵 2010-08-17 20:07 发表评论

Poj 2719 Faulty Odometer

田兵 — Tue, 17 Aug 2010 03:17:00 GMT

里程表逢9进1，大于的位置减1后就变成九进制表示的结果。

其中strtol返回十进制结果，strtol(str,0,base);返回将base进制的数转化成十进制后的值。

#include
#include
#include    
int main()
{
    char data[15]={0};
    while(scanf("%s",data)&&!(strcmp(data,"0")==0))
    {
                      int b,i,j;
                      char c[15];
                      strcpy(c,data);
                      j=strlen(data);
                      for(i=0; i                            if(data[i]>'4')data[i]-=1;
                        b=strtol(data,0,9);
                        printf("%s: %d\n",c,b);

    }
    return 0;
}

田兵 2010-08-17 11:17 发表评论

poj 1094 Sorting It All Out Folyd+dfs拓扑排序

田兵 — Tue, 17 Aug 2010 02:26:00 GMT

每接受一组关系，用Floyd算法计算传递闭包，

判断是否有矛盾：if(adj[i][j]&&adj[j][i])isInconsitent=true;

判断是否有关系未确定：if(adj[i][j]==0&&adj[j][i]==0&&(i!=j))issorted=false;

如果没有矛盾，所以关系确定，则说明已经sort好了，用dfs拓扑排序输出即可。

然后把未接受的数据接受完，刚开始直接跳出了。

#include<iostream>
#include<cstring>
#include<vector>
using namespace std;
bool adj[27][27]; int n,m;
bool visit[27],isInconsitent=false,issorted=true;
vector<int>vec;
void Floyd()
{
    int i,j,k;
    for(k=1; k<=n; k++)
        for(i=1; i<=n; i++)
            for(j=1; j<=n; j++)
                 adj[i][j]= (adj[i][j]||adj[i][k]&&adj[k][j]);
}

void check()
{
    isInconsitent=false,issorted=true;
    int i,j;
    for(i=1; i<=n; i++)
        for(j=1; j<=n; j++)
        {
            if(adj[i][j]&&adj[j][i])isInconsitent=true;
            if(adj[i][j]==0&&adj[j][i]==0&&(i!=j))issorted=false;
        }
}

void dfs(int i)
{
    visit[i]=true;
    for(int j=1; j<=n; j++)
        if(!visit[j]&&adj[i][j])
            dfs(j);
    vec.push_back(i);
}
int main()
{
    while(cin>>n>>m)
    {
        if(n==0&&m==0)break;
        memset(adj,0,sizeof adj);
        bool print=false;
        int i,j;
        char pre,last,less;
        for(i=1; i<=m; i++)
        {
            cin>>pre>>less>>last;
            adj[int(pre-'A'+1)][int(last-'A'+1)]=true;
            
            Floyd();

            check();
            
            if(isInconsitent)
            {
                cout<<"Inconsistency found after "<<i<<" relations."<<endl;
                print =true; break;
            }

            if(issorted)
            {
                cout<<"Sorted sequence determined after "<<i<<" relations: ";
                vec.clear();
                memset(visit,0,sizeof visit); 
                for(int t=1; t<=n; t++)
                    if(!visit[t])dfs(t); 
                while(vec.size())
                {
                    cout<<char(vec[vec.size()-1]+'A'-1);
                    vec.pop_back(); 
                }
                cout<<'.'<<endl;
                print=true; break;
            }
            
        }
        if(print==false)cout<<"Sorted sequence cannot be determined."<<endl;
        //accept the data left
        for(j=i+1; j<=m; j++)
            cin>>pre>>less>>last;
    }

    return 0;
}

田兵 2010-08-17 10:26 发表评论

poj 2531 Network Saboteur

田兵 — Sun, 15 Aug 2010 13:40:00 GMT

题意：
将一个图分为两个图A，B，ans最大，ans=Cij的和(i属于A，j属于B)

直接搜索1700+水过，f[i]==0,f[i]==1分别表示属于A和B，不知道0ms是怎么剪出来的

#include<iostream>
using namespace std;

int adj[21][21]={0};
bool f[21]={0};
int n,ans=0,max=0;

int cal()
{
    int sum=0;
    for(int i=1; i<=n; i++)
    for(int j=1; j<i; j++)
    {
           if(f[i]!=f[j])
           sum+=adj[i][j]; 
    }
    
    for(int i=2; i<=n; i++)
            if(f[1]!=f[i])
            return sum;
    return 0;
}
void search(int i)
{
     if(i>n)
     {
            int temp=cal();
            if(temp>ans)ans=temp;
            return ;
     }
     f[i]=1;
     search(i+1);
     f[i]=0;
     search(i+1);
}
int main()
{
    cin>>n;
    for(int i=1; i<=n; i++)
    for(int j=1; j<=n; j++)
            cin>>adj[i][j];
    
    search(1);
    
    cout<<ans<<endl;
    system("pause");
    return 0;
}

田兵 2010-08-15 21:40 发表评论

POJ 1141 Washing Clothes 背包问题

田兵 — Fri, 13 Aug 2010 09:00:00 GMT

背包问题：
对每种颜色的衣服求解背包问题，和即为结果。

对于第i中颜色，所有时间和为sum，pack(i)返回洗该种衣服所花的最少时间。
最少时间问题就是包容量为sum/2的01背包问题，sum-dp[sum/2]就是结果

#include
#include
#include
using namespace std;
int adj[105][105];
int dp[100005];
int pack(int i)
{
      memset(dp,0,sizeof dp);
      int sum=0,k,j;
      for(k=1; k<=adj[i][0]; k++)
               sum+=adj[i][k];
      int v=sum/2;
      for(k=1; k<=adj[i][0]; k++)
      for(j=v; j>=adj[i][k]; j--)
      {
               if(dp[j]                   dp[j]=dp[j-adj[i][k]]+adj[i][k];
      }
      
      return sum-dp[sum/2];
}

int main()
{
    int m,n,j,i,value;
    string str;
    while(cin>>m>>n,n||m)
    {
        memset(adj,0,sizeof adj);
        vector vec;
        for(i=1;i<=m; i++ )
               {
                      cin>>str;
                      vec.push_back(str);
               } 
                     
        for(i=1; i<=n; i++)
        {
               cin>> value>>str;
               for(j=0; j               {
                        if(str==vec[j])
                        {
                            adj[j][0]++;
                            adj[j][adj[j][0]]=value;
                        }
               } 
        }
        
        int ans=0;
        for(i=0; i                 ans+=pack(i);
        cout<                    
    }
    
    system("pause");
    return 0;
}

田兵 2010-08-13 17:00 发表评论

POJ 1276 Cash Machine

田兵 — Mon, 09 Aug 2010 15:52:00 GMT

79ms   Cash Machine

多重背包问题：
思想：
把多重背包问题转化成01背包，假设某件物品的数量上限是n,每件的体积是c，价值是w
那么可把该物品分成系数是1,2,4……，2^(k-1), 和n-2^k+1,（k的满足2^k 那么新的物品就是（体积，价值）（1*c，1*w），（2*c，2*w），（4*c，4*w），
            ……（2^k*c，2^k*w），（（n-2^k+1）*c，(n-2^k+1)*w）
    例如7可以分成系数为1,2，4。 13得到系数为1,2,4,6的物品。
   这样构造出来的物品和原来数量小于等于n的情况等同，即原来该物品可取的任意数量，可以由这些物品组合得到。
   如比如7那个例子，取6个物品，可由2,4得到。取3个物品可由，1,2得到。
   deal()就是完成这样的任务。

1
2 #include<iostream>
3 #include<algorithm>
4 #include<string.h>
5 using namespace std;
6 int c[10001]={0};
7 int dp[100001]={0};
8 int i,j,n,cash,npack,bill,num;
9
10 void deal(int num, int bill)
11 {
12     int k=0,j,t;
13     if(num==1){ npack++; c[npack]=bill; return ; }
14     if(num==2){ npack++; c[npack]=bill; npack++; c[npack]=bill; return ;}
15     for( k=1,j=2*k; 2*j-1<num; k++,j*=2)  //j==2^k
16                     ;
17     k--;
18     for(j=1,t=0; t<=k; j*=2,t++)
19     {
20              npack++;
21              c[npack]=j*bill;
22     }
23     npack++;
24     c[npack]=(num-j+1)*bill;
25 }
26
27 int main()
28 {
29
30     while(cin>>cash)
31     {
32        memset(dp,0,sizeof dp);
33        memset(c,0,sizeof c);
34        npack=0;
35        cin>>n;
36        for(i=1; i<=n; i++)
37                {
38                       cin>>num>>bill;
39                       if(num==0)continue;
40                       deal(num,bill);
41                }
42
43
44
45        dp[0]=1;
46
47        for(i=1; i<=npack; i++)
48        for(j=cash; j>=c[i]; j--)
49        {
50                    dp[j]= dp[j]||dp[j-c[i]];
51        }
52
53        for(j=cash; j>=0; j--)
54                   if(dp[j])
55                   {
56                            cout<<j<<endl;
57                            break;
58                   }
59     }
60     system("pause");
61     return 0;
62 }
63
64

田兵 2010-08-09 23:52 发表评论

POJ 3259 Wormholes

田兵 — Mon, 09 Aug 2010 11:38:00 GMT

Bellman_Ford 算法，

利用Bellman_Ford算法判断是否存在负回路。如果存在负回路就一定可以回到（不停的走这个回路，直到赚到的负值够与回去的值抵消的），如果不存在一定不可以回去。
用cin500+ms，用scanf 100+ms

#include<iostream>
#include<stdio.h>
using namespace std;
const int INF=0x0fffffff;
struct type
{
       int s,t;
       int len;
}edge[30000];

int n,m,w,nedge=0;
int d[505];
bool Bellman_Ford()
{
     bool isPossible=false;
     for(int j=1; j<=n; j++)d[j]=INF;
     d[1]=0;
     int u,v;
     bool f;
     for(int j=1; j<=n-1; j++)
     {
             f=0;
             for(int k=1; k<=nedge; k++)
             {
                     u=edge[k].s; v=edge[k].t;
                     if(d[v]>d[u]+edge[k].len)
                                 d[v]=d[u]+edge[k].len;
                     f=1;            
             }
             if(f==0)return false;
     }
  /*前面n-1次循环可以求出无负回路时的最短路径，再来一次如果可以更新就一定有负回路。/*
      for(int k=1; k<=nedge; k++)
             {
                     u=edge[k].s; v=edge[k].t;
                     if(d[v]>d[u]+edge[k].len)
                           {  d[v]=d[u]+edge[k].len; return 1;}
                                
             }
     
    // system("pause");
     return 0;
     
}

int main()
{
    int t;
    cin>>t;
    while(t--)
    {
              int i,j,u,v,value;
              scanf("%d %d %d",&n,&m,&w);//cin>>n>>m>>w;
              nedge=0;
              for(int i=1; i<=m; i++)
              {
                      scanf("%d %d %d",&u,&v,&value);//cin>>u>>v>>value;
                      ++nedge;
                      edge[nedge].s=u;
                      edge[nedge].t=v;
                      edge[nedge].len=value;
                      ++nedge;
                      edge[nedge].s=v;
                      edge[nedge].t=u;
                      edge[nedge].len=value;
              }     
                  
              for(int i=1; i<=w; i++)
              {
                      scanf("%d %d %d",&u,&v,&value);//cin>>u>>v>>value;
                      ++nedge;
                      edge[nedge].s=u;
                      edge[nedge].t=v;
                      edge[nedge].len=-value;
                      
                     /* ++nedge;
                      edge[nedge].s=v;
                      edge[nedge].t=u;
                      edge[nedge].len=-value;
                      */
              }
              
              if(Bellman_Ford())
                         cout<<"YES"<<endl;
              else       cout<<"NO"<<endl;
     
    }
    system("pause");
    return 0;
}

田兵 2010-08-09 19:38 发表评论

POJ 2485 Highways

田兵 — Mon, 09 Aug 2010 08:11:00 GMT

188MS

Kruskal算法，cin竟然超时，输入数据的时间比程序运行的时间要多的多
并查集每次用过之后忘记更新了： fa[x]=y;

#include<stdio.h>
#include<algorithm>
using namespace std;

struct node
{
       int s,t,len;
}arr[500*500+100];

int fa[500*500+100];
bool cmp(const node &a, const node & b)
{
     return a.len<b.len;
}

int find(int x)
{
    return ( fa[x]==x?x:fa[x]=find(fa[x]) ) ;
}

int main()
{
    int t,n,i,j,temp,nedge;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);// cin>>n;
        nedge=0;
        for(i=1; i<=n; i++)fa[i]=i;
        for(i=1; i<=n; i++)
        for(j=1; j<=n; j++)
        {
                 scanf("%d",&temp);//cin>>temp;
                 if(j>=i)continue;
                 nedge++;
                 arr[nedge].s=i;
                 arr[nedge].t=j;
                 arr[nedge].len=temp; 
        }
        sort(arr+1,arr+1+nedge,cmp);
        int ans=0;
        for(i=1,j=0; i<=nedge; i++)
        {
                 int x=find(arr[i].s),y=find(arr[i].t);
                 if(x!=y){
                  fa[x]=y;
                  ans=arr[i].len;   j++; 
                  if(j==n-1)break;
                   }          
        }
        printf("%d\n",ans);//cout<
    }
    system("pause");
    return 0;
}

田兵 2010-08-09 16:11 发表评论

poj 1007 DNA Sorting

田兵 — Mon, 09 Aug 2010 07:04:00 GMT

#include<iostream>
#include<string>
#include<algorithm>
using namespace std;
struct type
{
 string str;
 int cnt,i;
 }arr[105];

int cal(string &str)
{
    int cnt=0,i,j;
    for(i=0; i<str.size(); i++)
            for(j=i+1; j<str.size(); j++)
            {
                 if(str[j]<str[i])cnt++;   
            }
    return cnt;
}

bool cmp(const   type &a,const  type &b)
{
     if(a.cnt<b.cnt)return 1;
     if(a.cnt==b.cnt)return a.i<b.i;
     return 0;
}

int main()
{
    int n,m;
    cin>>n>>m;
    string str;
    for(int i=1; i<=m; i++)
            {
                       cin>>arr[i].str;
                       arr[i].i=i;
                       arr[i].cnt=cal(arr[i].str);
            }
       
    
    sort(arr+1,arr+1+m,cmp);
    
    for(int i=1; i<=m; i++)
    cout<<arr[i].str<<endl;
    
    //system("pause");
    return 0;
} 

田兵 2010-08-09 15:04 发表评论