POJ 3660 Cow Contest（传递闭包floyed算法）

╰+哭是因爲堅強的太久メ 2022-08-09 13:46 149阅读 0赞

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

Source

[USACO 2008 January Silver][]

题目大意：给定N个人，然后知道这N个人的两两对决的情况，前一个为胜者，问M个对决情况中最多能确定几个人的名次.

思路：想到了Floyd算法，可是怎么来关联还是不知所以然，网上查了资料后，才明白用Floy求传递闭包，只要该人的关系和其他人的关系确定，那么他的名次就是确定的。

Code

#include <iostream>
    #include <cstring>
    #include <algorithm>
    using namespace std;
    
    const int N=110;
    int mp[N][N];
    
    int main()
    {
        
        int n,m;
        while (cin>>n>>m)
        {
        
            int i,j,k;
            int u,v;
            memset(mp,0,sizeof(mp));
            for (i=0;i<m;i++)
            {
        
                cin>>u>>v;
                mp[u][v]=1;
            }
           for(k=1;k<=n;k++)
              for(i=1;i<=n;i++)
                  for(j=1;j<=n;j++)
                    if(mp[i][k]==1&&mp[k][j]==1)  //A和B有关，B和C有关，则Ac有关，不等号的传递性                     mp[i][j]=1;
            //Floyd();
            int ans=0;
            for (i=1;i<=n;i++)
            {
        
                for (j=1;j<=n;j++)
                {
        
                    if (j==i) continue;
                if (mp[i][j]==0&&mp[j][i]==0) break;
                }
                if (j>n) ans++;
            }
            cout<<ans<<endl;
        }
    }

[USACO 2008 January Silver]: http://poj.org/searchproblem?field=source&key=USACO+2008+January+Silver