POJ 3660 Floyd求传递闭包

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 5
4 3
4 2
3 2
1 2
2 5
Sample Output
2

题意就是：给出你你x奶牛能战胜y奶牛，问你能确定哪些奶牛的排名

一个传递闭包问题，实际上就是用FLoyd求出任意一个点，能到达那些的以及那些的能到达这个点，如果某一个点，符合能到达他的点的个数加上他能到达点的个数 == n-1（没有他自己），那么他的排名就能确立，不明白可以把样例画出来

代码如下：

#include<iostream>#include<cstring>#include<cstdio>#include<cmath>#include<vector>#include<string>#include<algorithm>using namespace std;const int MAX_V = 110;bool e[MAX_V][MAX_V];int N,M;void floyd(){    for(int k=1;k<=N;k++){//floyd中k必须要放在最外层        for(int i=1;i<=N;i++){            for(int j=1;j<=N;j++){                if(e[i][k] && e[k][j])                    e[i][j] = true;            }        }    }}int main(void){    while(scanf("%d %d",&N,&M) != EOF){        int x,y;        memset(e,false,sizeof(e));        for(int i=1;i<=M;i++){            scanf("%d %d",&x,&y);                e[x][y] = true;        }        floyd();        int res = 0;        for(int i=1;i<=N;i++){            int sum = 0;            for(int j=1;j<=N;j++){                if(i == j)  continue;                if(e[i][j] || e[j][i])                    sum++;            }            if(sum == N-1)  res++;        }        printf("%d\n",res);    }    return 0;}