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传送门
Cow Contest
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 24342 Accepted: 13539
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 5
4 3
4 2
3 2
1 2
2 5

Sample Output

题目大意

能知道哪些牛比哪些牛强，问可以确定排名的牛有多少。

解题思路

能确定排名，那么这头牛与其他牛的关系得是确定的，就利用形似floyd算法的方式找传递闭包。

AC代码

#include<cstdlib>
#include<cstring>
#include<cstdio>
using namespace std;const int maxn = 101;
bool f[maxn][maxn];
int n, m, ans = 0;void floyd(void){for(int k = 1; k <=n ; k++)for(int i = 1; i <=n ; i++)for(int j = 1; j <= n; j++)f[i][j]=(f[i][k]&&f[k][j]) ? 1 : f[i][j];
}int main(){memset(f,0,sizeof(f));scanf("%d%d", &n,&m);for(int i = 1, u, v; i <= m; i++){scanf("%d%d",&u,&v);f[u][v]=1;}floyd();for(int i = 1,cnt;  i <= n; i++){cnt = 0;for(int j = 1; j <= n; j++){cnt+=f[i][j]||f[j][i];}if(cnt == n-1)ans++;}printf("%d\n",ans);// system("pause");return 0;
}

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[传递闭包]POJ#3660 Cow Contest

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