POJ 3660 Cow Contest (闭包传递)

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7690		Accepted: 4288

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




刚才学了下闭包传递，简单来说就是用Floyd来求两点是否连通。这题里，如果一点的入度加上出度等于N-1，那么就可确定这点的等级，因为除了它自己之外的所有节点要么在它之前要么在它之后，不管它前面的节点后后面的节点怎么排序，都不会影响到它。

#include <iostream>
#include <cstdio>
using    namespace    std;

const    int    SIZE = 105;
int    N,M;
int    IN[SIZE],OUT[SIZE];
bool    G[SIZE][SIZE];

int    main(void)
{
    int    from,to;
    int    ans;

    while(scanf("%d%d",&N,&M) != EOF)
    {
        fill(&G[0][0],&G[N][N],false);
        for(int i = 0;i <= N;i ++)
            IN[i] = OUT[i] = 0;

        for(int i = 0;i < M;i ++)
        {
            scanf("%d%d",&from,&to);
            G[from][to] = true;
        }

        for(int k = 1;k <= N;k ++)
            for(int i = 1;i <= N;i ++)
                for(int j = 1;j <= N;j ++)
                    G[i][j] = G[i][j] || G[i][k] && G[k][j];
        for(int i = 1;i <= N;i ++)
            for(int j = 1;j <= N;j ++)
                if(G[i][j])
                {
                    IN[j] ++;
                    OUT[i] ++;
                }

        ans = 0;        
        for(int i = 1;i <= N;i ++)
            if(IN[i] + OUT[i] == N - 1)
                ans ++;
        printf("%d\n",ans);
    }

    return    0;
}