POJ 3660 Cow Contest(传递闭包)

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：给出n只奶牛，m个胜负关系，求有几只奶牛的排名可以确定？

题解：传递闭包什么的，听着很高级，其实非常滑稽，感觉自学啥的也就两分钟，其实也就是一个floyd的延伸
用floyd根据三头牛之间a-b，b-c的胜负关系来判断a-c的胜负关系，然后显然除了和自己以外，如果和其他奶牛的关系已经确定了，他的排名也就确定了
然后就写出来了

代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

int map[][],n,m;

int main()
{
    memset(map,-,sizeof(map));
    scanf("%d%d",&n,&m);
    int xx,yy;
    for(int i=;i<=m;i++)
    {
        scanf("%d%d",&xx,&yy);
        map[xx][yy]=;
        map[yy][xx]=;
    }
    for(int i=;i<=n;i++)
    {
        map[i][i]=;
    }
    for(int k=;k<=n;k++)
    {
        for(int i=;i<=n;i++)
        {
            for(int j=;j<=n;j++)
            {
                if(map[i][k]==&&map[k][j]==)
                {
                    map[i][j]=;
                }
                if(!map[i][k]&&!map[k][j])
                {
                    map[i][j]=;
                }
            }
        }
    }
    int ans=;
    for(int i=;i<=n;i++)
    {
        int flag=;
        for(int j=;j<=n;j++)
        {
            if(map[i][j]==-)
            {
                flag=;
            }
        }
        ans+=flag;
    }
    printf("%d\n",ans);
}