POJ 3660 Cow Contest

题目链接：http://poj.org/problem?id=3660

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10066		Accepted: 5682

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

2

题目大意：有n头牛，然后有m个条件 每个条件有两个整数a，b表示a能打败b  然后问你有多少头牛能够确定名次。
解题思路：考虑到能够确定名次的牛需要满足一个条件（打败他的牛的个数加上他打败的牛的个数为n-1）
AC代码：

#include <stdio.h>
#include <string.h>
int p[110][110];
int n,m;
void floyd()
{
    int i,j,k;
    for (k = 1; k <= n; k ++)
    {
        for (i = 1; i <= n; i ++)
        {
            for (j = 1; j <= n; j ++)
            {
                if (p[i][k] && p[k][j])  //间接相连也表示能够打败
                    p[i][j] = 1;
            }
        }
    }
}
int main ()
{
    int i,j,a,b;
    while (~scanf("%d%d",&n,&m))
    {
        memset(p,0,sizeof(p));

        for (i = 0; i < m; i ++)
        {
            scanf("%d%d",&a,&b);
            p[a][b] = 1;
        }
        floyd();
        int ans,sum = 0;
        for (i = 1; i <= n; i ++)
        {
            ans = 0;
            for (j = 1; j <= n; j ++)
            {
                ans += p[i][j];  //他打败的牛的个数
                ans += p[j][i];  //打败他的牛的个数
            }
            if (ans == n-1)
                sum ++;
        }
        printf("%d
",sum);
    }
    return 0;
}