H - Cow Contest

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本文介绍了一种利用Floyd算法解决牛群编程竞赛中精确确定牛技能排名的方法。通过分析给定的比赛结果，该算法能够找出那些可以明确其技能等级的参赛牛。

H - Cow Contest

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

Sample Output

题意：

John 要给他的牛排名，于是让他的牛两两决斗。输入给出n个牛，m次决斗，又给出决斗者A和B，A为胜者；要你求出通过这些决斗有几个牛能确定名次；

思路：

用Floyd算法，遍历每头牛，如果有这么一头牛，比他排名高的和比他排名第的总数为n-1，则这头牛排名已经确定；

代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int INF=99999999;
int tu[110][110];
int main()
{
    int n,m,a,b,num=0,sum=0;
    scanf("%d%d",&n,&m);
    
    memset(tu,0,sizeof(tu)); //全部清0；
    
    for(int i=1;i<=m;i++)
    {
        scanf("%d%d",&a,&b);
        tu[a][b]=1;
    }
    for(int k=1;k<=n;k++)        //floyd算法
        for(int j=1;j<=n;j++)
            for(int i=1;i<=n;i++)
                if(tu[j][k]&&tu[k][i])
                    tu[j][i]=1;
                
    for(int i=1;i<=n;i++)
    {
        sum=0;
        for(int j=1;j<=n;j++)
        {
            sum+=tu[j][i]+tu[i][j];
        }
        if(sum==n-1)
            num++;
    }
    printf("%d\n",num);
}