Cow Contest---牛的比赛

本文已参与「新人创作礼」活动，一起开启掘金创作之路。​

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 cowB.

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：

是给出m对牛的相互关系，求有多少个牛排名是确定的。

方法：floyed

如果一个牛和其余的牛关系都是确定的，那么这个牛的排名就是确定的了~

 这个题有个注意点，在ac上有显示~

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int d[110][110];
int a,b;
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        memset(d,0,sizeof(d));
        for(int i=1; i<=m; i++)
        {
            scanf("%d %d",&a,&b);
            d[a][b]=1;           ///有关系为1
        }
        for(int k=1; k<=n; k++)///1 ≤ A ≤ N; 1 ≤ B ≤ N; 
        {
            for(int i=1; i<=n; i++)///循环中不能从0~n-1只能1~n！！！
            {
                for(int j=1; j<=n; j++)
                {
                    d[i][j]=d[i][j]||(d[i][k]&&d[k][j]);///有关系
                }
            }
        }
        int ans=0;
        for(int i=1; i<=n; i++)
        {
            int sum=0;
            for(int j=1; j<=n; j++)
            {
                if(d[i][j]||d[j][i])
                    sum++;
            }
            if(sum==n-1)          ///和n-1个有关系，就能确定位置
                ans++;
        }
        printf("%d\n",ans);
    }
    return 0;
}