AYIT-2020 609第十六周周赛 A - 闭包传递

最新推荐文章于 2020-06-07 22:08:31 发布

原创最新推荐文章于 2020-06-07 22:08:31 发布 · 225 阅读

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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，每头牛拥有独特的技能评级。通过分析M轮两两对决的比赛结果，本文介绍了一种算法，用于确定能够精确排名的牛的数量。该算法利用了图论中的闭包传递概念，通过三重for循环进行松弛操作，最终确定每头牛与其他所有牛的关系，从而判断其排名是否可以确定。

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

有N头牛，评以N个等级，各不相同，先给出部分牛的等级的高低关系，问最多能确定多少头牛的等级
解题思路：一头牛的等级，当且仅当它与其它N-1头牛的关系确定时确定，于是我们可以将牛的等级关系看做一张图，然后进行适当的松弛操作，得到任意两点的关系，再对没一头牛进行检查即可

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int dp[1005][1005];


int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m)&&(n+m))
	{
		for(int i=1;i<=m;i++)
		{
			int x,y;
			scanf("%d%d",&x,&y);
			dp[x][y]=1;// x和y之间有关系
		}
		for(int k=1;k<=n;k++)//闭包传递三重for循环
		{
			for(int i=1;i<=n;i++)
			{
				for(int j=1;j<=n;j++)
				{
					dp[i][j]=dp[i][j]||(dp[i][k]&&dp[k][j]);//判断x和y之间是否有直接或间接关系
				}
			}
		}
		int s=0,cnt=0;
		for(int i=1;i<=n;i++)
		{
			s=0;
			for(int j=1;j<=n;j++)
			{
				if(dp[i][j]||dp[j][i])//(x和y)或(y和x)之间有关系
				s++;	
			}
			if(s==n-1) cnt++;//if(关系数=(n-1)头牛)--->确定的牛数
		}
		printf("%d\n",cnt);
	}
 }