Cow Contest POJ

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：n个人，m个关系，每个关系（a b）告诉你,a比b水平高，问你最后有多少人水平是可以确定的。
思路:刚开始一开这题意就莽拓扑排序，后来发现不对，（想着如果入队多个，说明无序，入队一个就有序，简直睿智）
其实是用floyd计算传递闭包，最后看对每个人是否关系都确定了。

（因为我图中只记录了，谁比谁等级高，所以闭包出来，也只有该点是否比其他点等级高，例如 maps【a】【b】，显然缺少一种别点比他高的情况，其实就是maps【b】【a】）

#include<cstdio>
#include<iostream>

int maps[105][105];

int n,m;
int main()
{
    scanf("%d%d",&n,&m);
    for(int j=1; j<=m; j++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        maps[u][v] = 1;
    }
    for(int k=1; k<=n; k++)
    {
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                maps[i][j] |= maps[i][k] && maps[k][j];
            }
        }
    }
    int num=n;
//    for(int i=1;i<=n;i++)
//    {
//        for(int j=1;j<=n;j++)
//        {
//            printf("%d  ",maps[i][j]);
//        }
//        puts("");
//    }
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            if(j == i)continue;
            if(!maps[i][j] && !maps[j][i])
            {
                num--;
                break;
            }
        }
    }
    printf("%d
",num);
}