POJ3660 Cow Contest floyd传递闭包

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

Sample Input

Sample Output

题意：

　　有n头牛比赛，m种比赛结果，最后问你一共有多少头牛的排名被确定了，其中如果a战胜b，b战胜c，则也可以说a战胜c，即可以传递胜负。求能确定排名的牛的数目。

思路：

　　如果一头牛被x头牛打败，打败y头牛，且x+y=n-1，则我们容易知道这头牛的排名就被确定了，所以我们只要将任何两头牛的胜负关系确定了，在遍历所有牛判断一下是否满足x+y=n-1，将满足这个条件的牛数目加起来就是所求解。如果出度+入度=顶点数-1，则能够确定其编号。

代码：

#include<stdio.h>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int map[][];
int n,m;
void floyd()
{
    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]==)//关系传递
                    map[i][j]=;
}
int main()
{
    cin>>n>>m;
    memset(map,,sizeof(map));
    for(int i=;i<m;i++)
    {
        int a,b;
        cin>>a>>b;
        map[a][b]=;//单向边
    }
    floyd();
    int ans=;
    for(int i=;i<=n;i++)
    {
        int sum=;
        for(int j=;j<=n;j++)
        {
            if((map[i][j]==||map[j][i]==)&&i!=j)
                sum++;//计算出度和入度
        }
        if(sum==n-)
            ans++;
    }
    cout<<ans;
}