POJ - 366 Cow Contest

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本文介绍了一种通过Floyd算法解决牛牛编程竞赛排名的方法。比赛中每头牛的能力值不同，通过比赛结果确定牛的精确排名。利用Floyd算法进行松弛处理，最终统计能与其他所有牛建立胜负关系的数量。

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Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13569		Accepted: 7556

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

题意：每头牛都有一个能力值，能力高的会打败能力低的牛，给出n头牛和m次两两比赛的结果，问如果按能力值从小到大排序，可以确定出几头牛的位次。

思路：能够确定出位次的牛，一定与其他n-1只牛都有固定的关系。因为牛数很少，所以用floyd松弛，然后遍历一遍，找出所有与其他牛都存在关系的牛的个数。

#include <iostream>
#include <cstdio>
#include <map>
#include <cstring>
#define ll long long
using namespace std;
int n,m;
int mp[200][200];
int main()
{
	while(cin>>n>>m)
	{
		memset(mp,0,sizeof(mp));
		while(m--)
		{
			int x,y;
			cin>>x>>y;
			mp[x][y]=1;
		}
		for(int k=1;k<=n;k++)
		{
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=n;j++)
				{
					if(mp[i][k]==1&&mp[k][j]==1)
					{
						mp[i][j]=1;
					}
				}
			}
		}
		int ans=0;
		for(int i=1;i<=n;i++)
		{
			int cnt=0;
			for(int j=1;j<=n;j++)
			{
				if(i==j)
				continue;
				if(mp[i][j]==1||mp[j][i]==1)
				cnt++;
			}
			if(cnt>=n-1)
			{
				ans++;
			}
		}
		printf("%d\n",ans);
	}
}