poj3660Cow Contest

这道题目是一道有向图的传递背包问题。。在有向图中，又是不必关心路径的长度，而只关心每两点间是否有通路，则可以用1和0表示“联通”“不联通”。则主要算法只需要把

if(d[i][j]>d[now][j])改成d[i][j]=d[i][j]||(d[i][k]&&d[k][j])即可。。这就叫做有向图的传递闭包。

题目如下：

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 6652 Accepted: 3625

Description

N (1 ≤ N ≤ 100) cows, conveniently numbered 1..N, are participating in a programming contest. As we all know, some cows code better than others. Each cow has a certain constant skill rating that is unique among the competitors.

The contest is conducted in several head-to-head rounds, each between two cows. If cow A has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤ B ≤ N; A ≠ B), then cow A will always beat cow B.

Farmer John is trying to rank the cows by skill level. Given a list the results of M (1 ≤ M ≤ 4,500) two-cow rounds, determine the number of cows whose ranks can be precisely determined from the results. It is guaranteed that the results of the rounds will not be contradictory.

Input

* Line 1: Two space-separated integers: N and M
* Lines 2..M+1: Each line contains two space-separated integers that describe the competitors and results (the first integer, A, is the winner) of a single round of competition: A and B

Output

* Line 1: A single integer representing the number of cows whose ranks can be determined
　

Sample Input

5 54 34 23 21 22 5

Sample Output

2

Source

USACO 2008 January Silver

#include<cstdio>#include<cstring>#define maxn 105int n,m;int d[maxn][maxn];void floyd(){    int i,j,k;    for(k=1;k<=n;k++)        for(i=1;i<=n;i++)          for(j=1;j<=n;j++)             d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);}int main(){    int i,j,u,v,sum;    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(d,0,sizeof(d));        for(i=1;i<=n;i++)            for(j=1;j<=n;j++)            {                if(i==j)                    d[i][j]=1;//自己到自己为联通的            }        sum=0;        for(i=1;i<=m;i++)        {            scanf("%d%d",&u,&v);            d[u][v]=1;        }        floyd();        for(i=1;i<=n;i++)       {            for(j=1;j<=n;j++)            {                if(d[i][j]||d[j][i])                    continue;                else                        break;            }            if(j==n+1) sum++;        }        printf("%d\n",sum);    }    return 0;}