C++博客-无声的月

RMQ POJ 3264 Balanced Lineup

Baron — Wed, 18 Aug 2010 08:09:00 GMT

Description

For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

Input

Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

Output

Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

Sample Input

Sample Output

6
3
0

Source

USACO 2007 January Silver

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
const int N=50005;
int maxd[N][20],mind[N][20],data[N],n;
void Init()
{
    int i,j;
    for(i=1;i<=n;i++)
    {
        maxd[i][0]=mind[i][0]=data[i];
    }
    for(j=1;j<=log((double)(n+1))/log(2.0);j++)
    {
        for(i=1;i+(1<<j)-1<=n;i++)
        {
            maxd[i][j]=max(maxd[i][j-1],maxd[i+(1<<(j-1))][j-1]);
            mind[i][j]=min(mind[i][j-1],mind[i+(1<<(j-1))][j-1]);
        }
    }
}
int getDif(int a,int b)
{
    int k=(int)(log((double)(b-a+1))/log(2.0));
    int max_=max(maxd[a][k],maxd[b-(1<<k)+1][k]);
    int min_=min(mind[a][k],mind[b-(1<<k)+1][k]);
    return max_ - min_;
}
int main()
{
    int i,m,a,b;
    scanf("%d%d",&n,&m);
    for(i=1;i<=n;i++)
    {
        scanf("%d",data+i);
    }
    Init();
    while(m--)
    {
        scanf("%d%d",&a,&b);
        cout<<getDif(a,b)<<endl;
    }

}

Baron 2010-08-18 16:09 发表评论

poj 2251 BFS 三维迷宫

Baron — Sun, 25 Jul 2010 03:14:00 GMT

#include<iostream>
#include<queue>
using namespace std;
/*
1 means it can go through
0 means it cann't go through
-1 means the exit point
2 means the starting point
*/
int maze[35][35][35],L,R,C,sx,sy,sz,dx,dy,dz,cnt;
struct point
{
    int x,y,z;
    int level;
};
bool v(int x,int y,int z)
{
    return x<=L&&y<=R&&z<=C;
}
int main()
{
    int i,j,k,x,y,z,nx,ny,nz,level,nlevel,px,py,pz;
    bool find;
    char ch;
    point p;
    memset(maze,0,sizeof(maze));

    while(cin>>L>>R>>C,L||R||C)
    {
        queue<point> q;
        find=false;
        for(i=1;i<=L;i++)
        {
            for(j=1;j<=R;j++)
            {
                for(k=1;k<=C;k++)
                {
                    cin>>ch;
                    switch(ch)
                    {
                        case 'S':maze[i][j][k]=-1;sx=i;sy=j;sz=k;break;
                        case '.':maze[i][j][k]=1;break;
                        case '#':maze[i][j][k]=0;break;
                        case 'E':maze[i][j][k]=2;dx=i;dy=j;dz=k;break;
                        default:
                            break;
                    }
                }
            }
        }

        p.x=sx;p.y=sy;p.z=sz;p.level=0;
        q.push(p);
      //  cout<
        while(!q.empty())
        {
            p=q.front();
            q.pop();
            x=p.x;y=p.y;z=p.z; level=p.level;
         //   cout<"<
            if(x==dx&&y==dy&&z==dz)
            {
                cnt=level;
                find=true;
                break;
            }
            for(i=0;i<6;i++)
            {
                switch(i)
                {
                    case 0:px=x-1;py=y;pz=z;break;
                    case 1:px=x+1;py=y;pz=z;break;
                    case 2:px=x;py=y+1;pz=z;break;
                    case 3:px=x;py=y-1;pz=z;break;
                    case 4:px=x;py=y;pz=z+1;break;
                    case 5:px=x;py=y;pz=z-1;break;
                }
                if(maze[px][py][pz]==1||maze[px][py][pz]==2)//don't forgot to let the last enter
                {
                    p.x=px;p.y=py;p.z=pz;p.level=level+1;
                    q.push(p);
                    maze[px][py][pz]=3;
                }
            }
        }
        if(find)
        {
            printf("Escaped in %d minute(s).\n",cnt);
        }
        else
        cout<<"Trapped!"<<endl;
    }
}

Dungeon Master

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6841		Accepted: 2718

Description

You are trapped in a 3D dungeon and need to find the quickest way out! The dungeon is composed of unit cubes which may or may not be filled with rock. It takes one minute to move one unit north, south, east, west, up or down. You cannot move diagonally and the maze is surrounded by solid rock on all sides.

Is an escape possible? If yes, how long will it take?

Input

The input consists of a number of dungeons. Each dungeon description starts with a line containing three integers L, R and C (all limited to 30 in size).
L is the number of levels making up the dungeon.
R and C are the number of rows and columns making up the plan of each level.
Then there will follow L blocks of R lines each containing C characters. Each character describes one cell of the dungeon. A cell full of rock is indicated by a '#' and empty cells are represented by a '.'. Your starting position is indicated by 'S' and the exit by the letter 'E'. There's a single blank line after each level. Input is terminated by three zeroes for L, R and C.

Output

Each maze generates one line of output. If it is possible to reach the exit, print a line of the form

Escaped in x minute(s).

where x is replaced by the shortest time it takes to escape.
If it is not possible to escape, print the line

Trapped!

Sample Input

3 4 5
S....
.###.
.##..
###.#
#####
#####
##.##
##...
#####
#####
#.###
####E
1 3 3
S##
#E#
###
0 0 0

Sample Output

Escaped in 11 minute(s).
Trapped!

Source

Ulm Local 1997

Baron 2010-07-25 11:14 发表评论

POJ 1459 最大流 Edmonds-Karp算法

Baron — Wed, 21 Jul 2010 02:55:00 GMT

Power Network

Time Limit: 2000MS		Memory Limit: 32768K
Total Submissions: 10383		Accepted: 5691

Description

A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= p_max(u) of power, may consume an amount 0 <= c(u) <= min(s(u),c_max(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= l_max(u,v) of power delivered by u to v. Let Con=Σ_uc(u) be the power consumed in the net. The problem is to compute the maximum value of Con.

An example is in figure 1. The label x/y of power station u shows that p(u)=x and p_max(u)=y. The label x/y of consumer u shows that c(u)=x and c_max(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and l_max(u,v)=y. The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.

Input

There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets (u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of l_max(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of p_max(u). The data set ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of c_max(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can occur freely in input. Input data terminate with an end of file and are correct.

Output

For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.

Sample Input

2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20
7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7
(3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5
(0)5 (1)2 (3)2 (4)1 (5)4

Sample Output

15
6

Hint

The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second data set encodes the network from figure 1.

Source

1 #include<iostream>
2 #include<queue>
3 #include<algorithm>
4 #include<string>
5 #include<cstdlib>
6 using namespace std;
7 const int M=105;
8 const int INF=1000000;
9 int cap[M][M],flow[M][M],p[M],a[M];
10 int n,np,nc,m,s,t,i,j,k,u,v,w,ans;
11 queue<int> q;
12 char str[100];
13 int main()
14 {
15     while(cin>>n>>np>>nc>>m)
16     {
17         memset(flow,0,sizeof(flow));
18         memset(cap,0,sizeof(cap));
19         ans=0;
20         s=n;
21         t=n+1;
22
23         for(i=0;i<m;i++)
24         {
25             scanf("%s",str);
26             sscanf(str,"(%d,%d)%d",&u,&v,&w);
27             cap[u][v]=w;
28         }
29         for(i=0;i<np;i++)
30         {
31             scanf("%s",str);
32             sscanf(str,"(%d)%d",&v,&w);
33             cap[s][v]=w;
34         }
35         for(i=0;i<nc;i++)
36         {
37             scanf("%s",str);
38             sscanf(str,"(%d)%d",&u,&w);
39             cap[u][t]=w;
40         }
41         while(true)
42         {
43             memset(a,0,sizeof(a));
44             a[s]=INF;
45             q.push(s);
46             while(!q.empty())
47             {
48                 u=q.front();
49                 q.pop();
50                 for(v=0;v<=t;v++)
51                 {
52                     if(!a[v]&&cap[u][v]>flow[u][v])
53                     {
54                         p[v]=u;
55                         q.push(v);
56                         a[v]=min(a[u],cap[u][v]-flow[u][v]);
57                     }
58                 }
59             }
60
61             if(a[t]==0)
62             {
63                 break;
64             }
65
66             for(u=t;u!=s;u=p[u])
67             {
68                 flow[p[u]][u]+=a[t];
69                 flow[u][p[u]]-=a[t];
70             }
71             ans+=a[t];
72         }
73         cout<<ans<<endl;
74     }
75 }
76
77
78
79

Coming from:http://hi.baidu.com/%8E%E1%D0%B3/blog/item/6fddc4c9c56ff512be09e635.html

Baron 2010-07-21 10:55 发表评论

a^b mod n 求最大公约数，及其扩展

Baron — Wed, 23 Jun 2010 03:43:00 GMT

a^b mod n

1long long  ModExp(long long  a, long long b, long long n)
2{
3   long long result=1, y=a;
4   while (b)
5  {
6     if (b%2 == 1)
7       result = result * y % n;
8     y = y * y % n;
9     b/=2;
10   }
11   return result;
12}
13

求最大公约数，及其扩展

1 int Euclid(int a,int b)//a>=b,求a,b的最大公约数
2 {
3     if(b==0)
4     return a;
5     else
6     return Euclid(b,a%b);
7 }
8 int EuclidExtension(int a,int b,int &x,int &y)//a>=b ，求d=ax+by的x,y，其中d为两者的最大公约数
9 {
10
11     if(b==0)
12     {
13         x=1;
14         y=0;
15         return 1;
16     }
17     else
18     {
19         int x1,y1;
20         int d1=EuclidExtension(b,a%b,x1,y1);
21         x=y1;
22         y=x1-((int)(a/b))*y1;
23         return d1;
24     }
25 }

Baron 2010-06-23 11:43 发表评论