poj 3660 Cow Contest（传递闭包floyed算法）

最新推荐文章于 2024-05-09 15:22:54 发布

elbadaernu

最新推荐文章于 2024-05-09 15:22:54 发布

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本文介绍了一个简单的传递闭包Floyd算法实现，用于解决牛群之间的技能排名问题。通过输入每场竞赛的结果，算法能够确定哪些牛的排名可以被精确地识别。

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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

5 5
4 3
4 2
3 2
1 2
2 5
Sample Output

大致题意：有多组数据，每组数据首先输入两个数n，m表示有n头牛，有m组对应关系，
接下来有m行，每行输入x，y，表示编号x的牛排名在y的前面，问有多少头牛的关系是确定的。

思路：简单的传递闭包floyed算法

代码如下

#include<iostream>
#include<algorithm>
#include<string>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<cstring>
using namespace std;
const int MAXN=110;
int win[MAXN][MAXN];

int main()
{
   int n,m;
   while(scanf("%d%d",&n,&m)!=EOF)
   {
     memset(win,0,sizeof(win));
    int x,y;
    while(m--)
    {
        scanf("%d%d",&x,&y);
        win[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(win[i][k]&&win[k][j])
          win[i][j]=1;

      int ans=n;//假设n头牛的关系都确定了
      int j;
      for(int i=1;i<=n;i++)
      {
        for(j=1;j<=n;j++)
        {
            if(i==j) continue;
            if(win[i][j]==0&&win[j][i]==0) break;//i和j之间关系不确定，结束循环
          }
        if(j<=n) ans--; //该牛关系不确定，数量减一
      }   
        printf("%d\n",ans);   
   }
   return 0;
}