（最短路）Cow Contest

最新推荐文章于 2022-12-30 11:11:43 发布

zufe_cst

最新推荐文章于 2022-12-30 11:11:43 发布

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分类专栏：最短路

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在一场涉及N头编号为1至N的牛的编程竞赛中，每头牛都有独特的技能等级。通过分析M轮两两对决的比赛结果，本文介绍了一个算法，用于确定能精确排名的牛的数量。

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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

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include<queue>
using namespace std;
const int maxn=101;
const int inf=0x3f3f3f3f;
int edge[maxn][maxn],n,t,m;
int main()
{
	int i,j,k;
	while(~scanf("%d%d",&n,&m))
	{
		int a,b;
		memset(edge,0,sizeof(edge));
		for(i=1;i<=m;i++) 
		{
			scanf("%d%d",&a,&b);
			edge[a][b]=1;
		}
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=n;j++)
			{
				for(k=1;k<=n;k++)
				{
					if(edge[j][i]==1&&edge[i][k]==1) edge[j][k]=1;
				}
			}
		}
		int ans=0;
		for(i=1;i<=n;i++)
		{
			int num=0;
			for(j=1;j<=n;j++)
			{
				if(edge[i][j]==1) num++;
				if(edge[j][i]==1) num++;
			}
			if(num==n-1) ans++;
		}
		printf("%d\n",ans);
	}
}