poj 3660(Floyd求传递闭包)

最新推荐文章于 2020-04-20 00:37:46 发布

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本文详细阐述了如何通过已知的牛与牛之间的比赛结果来确定所有牛的排名顺序，运用传递闭包理论解决实际问题。

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 9317		Accepted: 5249

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：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。

解题思路：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。

可以利用传递闭包，将某头牛到其它牛的"可达性"求出。

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

const int maxn = 105;
int n,m,map[maxn][maxn];

void floyd()
{
	for(int k = 1; k <= n; k++)
		for(int i = 1; i <= n; i++)
			for(int j = 1; j <= n; j++)
				map[i][j] = map[i][j] || (map[i][k] && map[k][j]);
}

int main()
{
	int u,v;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		memset(map,0,sizeof(map));
		for(int i = 1; i <= m; i++)
		{
			scanf("%d%d",&u,&v);
			map[u][v] = 1;
		}
		floyd();
		int ans = 0,tmp;
		for(int i = 1; i <= n; i++)
		{
			tmp = 0;
			for(int j = 1; j <= n; j++)
				if(i != j && (map[i][j] || map[j][i]))
					tmp++;
			if(tmp == n-1) ans++;
		}
		printf("%d\n",ans);
	}
	return 0;
}