C++博客-Dreams-随笔分类-DFS

hdu 2437 Jerboas

DreamSky — Wed, 13 May 2009 12:17:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=2437

//1364284 2009-05-13 20:11:33 Accepted 2437 187MS 4444K 1362 B C++ no way
#include <iostream>
#include <vector>
using namespace std;

const int N = 1004;

typedef struct
{
    int vex;
    int cost;
}node;

vector<node> vv[N];

int ans, e;
char path[N];
int num[N][N];//num[i][j]表示走到i结点花费mod K 的最小值
int n, m, s, k;
int pcnt;

void init()
{
    for(int i = 1; i <= n; i++)
        vv[i].clear();
    ans = e = -1;
    memset(num, -1, sizeof(num));
}

void make_graph()
{
    int a, b, cost;
    node cc;

    for(int i = 0; i < m; i++)
    {
        scanf("%d%d%d", &a, &b, &cost);

        cc.vex = b; cc.cost = cost;
        vv[a].push_back(cc);
       // follows the tunnels you can not go back to the starting burrow.
        //表示单向图
       // cc.vex = a;
        //vv[b].push_back(cc);
    }
}

void dfs(int v,int costs)
{
    int i,j,p,t;
    if(path[v] =='P' && costs % k == 0 && (ans ==-1 || ans > costs || (ans == costs && v < e)))
    {
        ans = costs;
        e = v;
    }
    j = vv[v].size();
    if(!j)
        return ;
    for(i=0;i<j;i++)
    {
        p = vv[v][i].vex;
        t = vv[v][i].cost + costs;
        if(num[p][t%k] == -1 || num[p][t%k] > t)
        {
            num[p][t%k] = t;
            dfs(p,t);
        }
    }
}

int main()
{
    int T,i, kk;
    scanf("%d", &T);
    for(i = 1; i <= T; i++)
    {
        scanf("%d%d%d%d", &n, &m, &s, &k);
        getchar();
        scanf("%s", path + 1);
        init();
        make_graph();
        dfs(s,0 );
        cout<<"Case "<<i<<": "<<ans<<" "<<e<<endl;
    }
    return 0;
}

DreamSky 2009-05-13 20:17 发表评论

hdu 1016 Prime Ring Problem

DreamSky — Thu, 16 Apr 2009 08:51:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1016

//1276949 2009-04-16 16:47:42 Accepted 1016 484MS 268K 918 B C++ no way
#include<iostream>
using namespace std;
int n,t;
int prime[38] = {0,0,2,3,0,5,0,7,0,0,0,11,0,13,0,0,0,17,0,19,0,0,0,23,0,0,0,0,0,29,0,31,0,0,0,0,0,37};
bool used[20];//标记i是否用过
int pre_num[20];
void dfs(int v,int num)
{
    int i;
    if(num == n)
    {
        if(prime[v+1] !=0 )
        {
            cout<<"1";
            for(i=2;i<t;i++)
                cout<<" "<<pre_num[i];
            cout<<endl;
        }
        return ;
    }
    else
    {
        for(i=1;i<=n;i++)
        {
            if(used[i] == false && prime[i+v] != 0 )
            {
                used[i] = true;
                pre_num[t++] = i;
                dfs(i,num+1);
                used[i] = false;
                t--;
            }
        }
    }
}
int main()
{
    int i,cas=1;
    while(cin>>n)
    {

        cout<<"Case "<<cas++<<":"<<endl;

        for(i=1;i<=n;i++)
            used[i] = false;
        used[1] = true;
        t = 1;
        pre_num[t++]=1;
        dfs(1,1);

        cout<<endl;
    }
    return 0;
}

DreamSky 2009-04-16 16:51 发表评论

hutc 1035 编辑距离问题

DreamSky — Thu, 16 Apr 2009 06:37:00 GMT

http://172.20.16.106/JudgeOnline/showproblem?problem_id=1035
http://acm.nankai.edu.cn/p1131.html
DP代码

#include<iostream>
#include<string>
using namespace std;
const int N = 2001;//用1001 Runtime Error，不要开得太小
int dp[N][N];//dp[i][j]表示串s的前i个字符匹配到串t的前j个字符的最小步数
int lens,lent;
char s[N],t[N];

int main()
{
    int i,j;
    while(scanf("%s%s",s,t)!=EOF)
    {
        lens = strlen(s);
        lent = strlen(t);

        for(i=0;i<=lens;i++)
            dp[i][0] = i;
        for(j=0;j<=lent;j++)
            dp[0][j] = j;//变换到第j个字符最多用j步

        for(i=1;i<=lens;i++)
        {
            for(j=1;j<=lent;j++)//目标串指针
            {
                if(s[i-1] == t[j-1])
                    dp[i][j] = dp[i-1][j-1];
                else //更改
                    dp[i][j] = dp[i-1][j-1] + 1;

                if(dp[i][j] > dp[i-1][j] + 1)//删除第i个
                    dp[i][j] = dp[i-1][j] + 1;

                if(dp[i][j] > dp[i][j-1] + 1)//插入第i个
                    dp[i][j] = dp[i][j-1] + 1;
            }
        }
        printf("%d\n",dp[lens][lent]);
    }
    return 0;
}

DFS代码（在南开JudgeOnline超时）

#include<iostream>
#include<string>
using namespace std;
const int M = 10000000;
const int N = 2001;//用1001 Runtime Error，不要开得太小
int dp[N][N];
int lens,lent;
char s[N],t[N];
int dfs(int i,int j)
{
    int p,q,r,mins=M;

    if(i == lens && j == lent )
        return 0;
    else if(i == lens && j != lent )
        return lent - j;
    else if(i != lens && j == lent )
        return lens - i;

    if(dp[i][j] != M) //做过的没有必要再做
        return dp[i][j];

    if(s[i] == t[j])
        mins = dfs(i+1,j+1);//不用改变
    else
    {
        p = dfs(i+1,j+1) + 1;//修改
        q = dfs(i+1,j) + 1;//删除
        r = dfs(i,j+1) + 1;//插入
        if(p > q)
            mins = q;
        else
            mins = p;
        if(mins > r)
            mins = r;
    }
    dp[i][j] = mins;
    return mins;
}

int main()
{
    int i,j;
    while(scanf("%s%s",s,t)!=EOF)
    {
        lens = strlen(s);
        lent = strlen(t);
        for(i=0;i<lens;i++)
            for(j=0;j<lent;j++)
                dp[i][j] = M;
        dfs(0,0);
        printf("%d\n",dp[0][0]);
    }
    return 0;
}

DreamSky 2009-04-16 14:37 发表评论

hdu 2141 Can you find it?

DreamSky — Sat, 11 Apr 2009 11:56:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=2141

#include<iostream>
#include<algorithm>
using namespace std;
__int64 a[505],b[505],c[505];
__int64 temp[250005];

bool bsearch(__int64 v,int low,int high) //二分查找
{
    int mid;
    while(low <= high)
    {
        mid = (high + low)/2;

        if(temp[mid] == v)
            return true;
        if( temp[mid] < v)
            low = mid+1;
        else
            high = mid-1;
    }
    return false;
}

int partition(__int64 r[],int i,int j)
{
    __int64 pivot = r[i];
    while(i<j)
    {
        while(i<j && r[j] >= pivot)
            j--;
        if(i<j)
            r[i++] = r[j];
        while(i<j && r[i] <= pivot)
            i++;
        if(i<j)
            r[j--] = r[i];
    }
    r[i] = pivot;
    return i;
}

void quickSort(__int64 r[],int low,int high) //快排
{
    int pivotpos;
    if(low < high)
    {
        pivotpos = partition(r,low,high);
        quickSort(r , low , pivotpos-1);
        quickSort(r , pivotpos+1 , high);
    }
}

int main()
{
    __int64 lena,lenb,lenc,cas=1;

    while(scanf("%I64d%I64d%I64d",&lena,&lenb,&lenc)!=EOF)
    {
        __int64 i,j,n;

        for(i=0 ; i < lena ; i++)

            scanf("%I64d",&a[i]);

        for(i=0 ; i < lenb ; i++)

            scanf("%I64d",&b[i]);

        for(i=0 ; i < lenc ; i++)

            scanf("%I64d",&c[i]);

        __int64 k=0;

        for(i=0 ; i< lena; i++)

            for(j=0 ; j< lenb; j++)

                 temp[k++] = a[i] + b[j];

        //sort(temp+0,temp+k); //,comp
        quickSort(temp,0,k-1);
        printf("Case %I64d:\n",cas++);

        scanf("%I64d",&n);
        while(n--)
        {
            int ok = 0;
            __int64 sum;
            scanf("%I64d",&sum);

            for(i=0; ok==0 && i<lenc ; i++)
            {
                if( bsearch( sum - c[i] ,0,k-1) )//if(binary_search(temp+0,temp+k,sum-c[i]))
                    ok = 1;
            }
            if(ok == 1)
                printf("YES\n");
            else
                printf("NO\n");
        }
    }
    return 0;
}

DreamSky 2009-04-11 19:56 发表评论

hdu 1010 Tempter of the Bone

DreamSky — Sat, 11 Apr 2009 11:26:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1010

#include<iostream>
using namespace std;
char gra[8][8];
bool mark[8][8];
int m,n,t,ok;
int si,sj,ei,ej;
int dir[4][2]={-1,0,1,0,0,-1,0,1};

int fun(int a,int b)
{
    if(a>b)
        return a-b;
    else
        return b-a;
}

void dfs(int i,int j,int step)
{
    int k,p,q;
    if(ok == 1)
        return ;
    if(step == t)
    {
        if(i == ei && j == ej)
            ok = 1;
        return ;
    }
    k = fun(i,ei) + fun(j,ej) ;
    if(step + k > t) //最快的时间都不能走到，肯定完蛋了
        return ;
    if(k % 2 != (t-step) % 2)//if((t - step - k) & 1) //奇偶剪枝
        return ;
    /**//*
       1 1 1 1 1 1
       1 0 1 1 1 1
       1 1 1 1 1 1
       1 1 1 1 1 1
       1 1 1 1 0 1
       两个0之间的坐标之差为6，是偶数，只能走偶数步才能到达
    */
    for(k=0;k<4;k++)
    {
       p = i + dir[k][0];
       q = j + dir[k][1];
       if(p>=0 && p<m && q>=0 && q<n && gra[p][q]!='X' && mark[p][q] == 0)
       {
            mark[p][q] = 1;
            dfs(p,q,step+1);
            mark[p][q] = 0;
       }
    }
}
int main()
{
    while(cin>>m>>n>>t && m+n+t)
    {
        int i,j,walls=0;
        for(i=0;i<m;i++)
            for(j=0;j<n;j++)
            {
                cin>>gra[i][j];
                if(gra[i][j] == 'S')
                {
                    si = i;
                    sj = j;
                }
                else if(gra[i][j] == 'D')
                {
                    ei = i;
                    ej = j;
                }
                else if(gra[i][j] == 'X')
                    walls++;
                mark[i][j] = 0;
            }
        if(n*m-walls<=t)
        {
           cout<<"NO"<<endl;
           continue;
        }
        ok = 0;
        mark[si][sj] = 1;
        dfs(si,sj,0);
        if(ok == 1)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}

DreamSky 2009-04-11 19:26 发表评论

hdu 1298 T9

DreamSky — Fri, 10 Apr 2009 08:03:00 GMT

摘要: http://acm.hdu.edu.cn/showproblem.php?pid=1298 //1254514 2009-04-10 15:46:40 Accepted 1298 0MS 600K 2448 B C++ no way #include阅读全文

DreamSky 2009-04-10 16:03 发表评论

hdu 1142 A Walk Through the Forest

DreamSky — Fri, 03 Apr 2009 12:27:00 GMT

http://acm.hdu.edu.cn/showproblem.php?pid=1142

#include<iostream>
using namespace std;
int n;//十字路口数
int map[1001][1001];
int dist[1001],dp[1001];
void dijkstra(int v)//迪杰斯特拉算法
{
    int i,j,mins,index;
    int *s = new int[n+1];
    for(i=1;i<=n;i++)
    {
        dist[i] = map[i][v];
        s[i] = 0;
    }
    dist[v] = 0;
    s[v] = 1;
    for(i=1;i<n;i++)
    {
        mins = 2000000;
        for(j=1;j<=n;j++)
        {
            if(s[j]==0 && dist[j]<mins)
            {
                mins = dist[j];
                index = j;
            }
        }
        if(mins == 2000000)
            break;
        s[index] = 1;
        for(j=1;j<=n;j++)
        {
            if(s[j]==0 && dist[j]>dist[index]+map[j][index])
                dist[j] = dist[index]+map[j][index];
        }
    }
}

int dfs(int v)//记忆法深搜
{
    if(dp[v] != -1)
        return dp[v];
    if(v == 2)
        return 1;
    int i,temp,sum=0;
    for(i=1;i<=n;i++)
    {
        if(map[v][i]!=2000000 && dist[v] > dist[i])//有路相通而且要去的i点到终点站的距离要比v到终点站的距离小
        {
            temp = dfs(i);
            sum += temp;
        }
    }
    dp[v] = sum;
    return sum;
}

int main()
{
    while(cin>>n && n)
    {
        int i,j,d,m;
        cin>>m;
        for(i=1;i<=n;i++)
        {
            dp[i] = -1;
            for(j=1;j<=n;j++)
                map[i][j] = 2000000;
        }
        while(m--)
        {
            scanf("%d%d%d",&i,&j,&d);
            map[i][j] = map[j][i] = d;//无向图
        }
        //求出各点到终点站的最短距离
        dijkstra(2);//2为终点站
        dfs(1);//从1出发
        cout<<dp[1]<<endl;
    }
    return 0;
}

DreamSky 2009-04-03 20:27 发表评论