poj-3660 cow contest

最新推荐文章于 2022-03-14 21:26:15 发布

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本文探讨了如何通过已知的奶牛编程竞赛结果来确定奶牛的排名。利用传递闭包算法解决该问题，重点在于计算每只奶牛能打败的其他奶牛数量，从而确定排名。

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Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 6098		Accepted: 3302

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

题目的意思：有很多奶牛，等级有高有低，等级高的可以打得过等级低的，让你通过已知的输入求得有哪些奶牛的排名可以确定。

分析：这里只要求出能够打败的和不能打败的个数有n-1个那么这个奶牛的排名就可以被确定。

通过这题学了一个新东西——“传递闭包”，是从网上搜到的

在此摘抄下来：

-------------------------

今天有人提到了传递闭包，我简单说说吧。
所谓传递性，可以这样理解：对于一个节点i，如果j能到i，i能到k，那么j就能到k。求传递闭包，就是把图中所有满足这样传递性的节点都弄出来，计算完成后，我们也就知道任意两个节点之间是否相连。
传递闭包的计算过程一般可以用Warshell算法描述：
For 每个节点i Do
    For 每个节点j Do
    If j能到i Then
        For 每个节点k Do
        a[j, k] := a[j, k] Or ( a[j, i] And a[ i, k] )
其中a数组为布尔数组，用来描述两个节点是否相连，可以看做一个无权图的邻接矩阵。可以看到，算法过程跟Floyd很相似，三重循环，枚举每个中间节点。只不过传递闭包只需要求出两个节点是否相连，而不用求其间的最短路径长。

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代码有点类似于floyd（）算法，判断语句注意顺序

#include<iostream>
#include<string>
using namespace std;
bool map[500][500];
void f(int n)
{
   int i, j, k;
   for(k = 1; k <= n; k++)
     for(i = 1; i <= n; i++)
        for(j = 1; j <= n; j++)
          if((map[i][k] && map[k][j]))  
              map[i][j] = true;
}
int main()
{
	int sum,i,num;
	int a,b,m,n,j;
	memset(map,false,sizeof(map));
	sum=0;
	cin>>n>>m;
	while(m--)
	{
		scanf("%d %d",&a,&b);
		map[a][b]=true;
	}
	f(n);
	for(i=1;i<=n;i++)
	{
		num=0;
		for(j=1;j<=n;j++)
		{
		
			if(map[i][j]||map[j][i])//找到他能打得赢的和打不赢的个数
				num++;
		}
		if(num==n-1)
		{
			sum++;
		}
	}
	cout<<sum<<endl;
}