POJ3660 Cow Contest —— Floyd 传递闭包

题目链接：http://poj.org/problem?id=3660

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13085		Accepted: 7289

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

2

Source

USACO 2008 January Silver

 

 

 

题解：

1.建图：如果A>B，则A—>B建一条边（B—>A也可以，但只能规定方向地建一条边）。

2.用Floyd求出传递闭包。

3.对于当前点X，如果与剩下的n-1个点都有联系，那么X的位置是可以确定的。

 

 

 

 

代码如下：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 #define rep(i,a,n) for(int (i) = a; (i)<=(n); (i)++)
 #define ms(a,b) memset((a),(b),sizeof((a)))
 using namespace std;
 typedef long long LL;
 const double EPS = 1e-;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e2+;

 int n, m;
 bool gra[MAXN][MAXN];

 int main()
 {
     while(scanf("%d%d", &n,&m)!=EOF)
     {
         memset(gra, false, sizeof(gra));
         for(int i = ; i<=m; i++)
         {
             int u, v;
             scanf("%d%d", &u,&v);
             gra[u][v] = true;
         }

         for(int k = ; k<=n; k++)   //求传递闭包
             for(int i = ; i<=n; i++)
                 for(int j = ; j<=n; j++)
                         gra[i][j] = gra[i][j] || (gra[i][k]&&gra[k][j]);

         int ans = ;
         for(int i = ; i<=n; i++)
         {
             int cnt = ;
             for(int j = ; j<=n; j++)
                 if( gra[i][j] || gra[j][i] )
                     cnt++;
             if(cnt==n-)    //i与剩下的n-1个数都能确定关系，则i的位置确定
                 ans++;
         }

         printf("%d\n", ans);
     }
 }