【POJ3660】【传递闭包】【能确定胜负的含义 +- = n-1】

本文描述了一场编程竞赛，其中N头编号为1到N的奶牛进行对决。每头奶牛都有独一无二的技能等级，奶牛之间的胜负由技能等级决定。根据M次对决的结果，需要确定多少头奶牛的排名可以精确得出。输入包括奶牛数量N、对决次数M以及每次对决的胜者，输出是可确定排名的奶牛数量。

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Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8371		Accepted: 4719

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 Input

Sample Output

Source

USACO 2008 January Silver

#include <iostream>
#include <cstring>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <string>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;
    
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp push_back

int N,M;
const int maxn = 110;
int mp[maxn][maxn];
void floyd()
{
	rep(k,1,N+1) rep(i,1,N+1) rep(j,1,N+1)
	{
		mp[i][j] = mp[i][j] || (mp[i][k] && mp[k][j]);
	}

}
int main()
{
    while(scanf("%d%d",&N,&M) != EOF)
	{
		memset(mp,0,sizeof(mp));
		rep(i,0,M) 
		{
			int u,v;
			scanf("%d%d",&u,&v);
			mp[u][v] = 1;
		}
		floyd();
		int ans = 0;
		rep(i,1,N + 1)
		{
			int cnt = 0;
			rep(j,1,N + 1)
			{
				if(i == j) continue;
				if(mp[i][j] || mp[j][i])
				{
					cnt ++;
				}
			}
		
			if(cnt == N-1) ans++;
		}
		printf("%d\n",ans);
	}
	
    return 0;
}