poj3660——Cow Contest

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

一开始以为是拓扑排序 百度了发现是用floyd求传递闭包，也就是先求出所有点之间的通路，然后如果一个点和其他的点都有通路的话，他的位置就能确定了

代码：

#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int N=105;
int g[N][N];
int n,m;
int u,v;
int main(void){
    scanf("%d%d",&n,&m);
    memset(g,0,sizeof(g));
    while(m--){
        scanf("%d%d",&u,&v);
        g[u][v]=1;
    }
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                if(g[i][k] && g[k][j]){
                    g[i][j]=1;
                }
            }
        }
    }
    int ans=0;
    int j;
    for(int i=1;i<=n;i++){
        for(j=1;j<=n;j++){
            if(i==j){
                continue;
            }
            if(!g[i][j] && !g[j][i]){
                break;
            }
        }
        if(j>n){
            ans++;
        }
    }
    printf("%d\n",ans);
    return 0;
}