Cow Contest(Poj 3660)

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

传递闭包，利用floyd-warshall算法解决

//============================================================================
// Name        : test.cpp
// Author      : Qihan
// Version     :
// Copyright   : Your copyright notice
// Description : Hello World in C++, Ansi-style
//============================================================================

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
using namespace std;
typedef long long int LLI;
typedef pair<LLI,LLI> PII;
#define Lowbit(x) (x & (-x))
const int inf = 1000000000;
const int maxn = (2048 + 10);

bool vis[maxn][maxn];

int main() {
//    freopen("/home/qihan/Documents/in","r",stdin);
    int n,m;
    memset(vis,false,sizeof(vis));
    scanf("%d%d",&n,&m);
    for(int i = 1; i <= n; i ++) {
        vis[i][i] = true;
    }
    for(int i = 1; i <= m; i ++) {
        int x,y;
        scanf("%d%d",&x,&y);
        vis[x][y] = true;
    }
    for(int k = 1; k <= n; k ++) {
        for(int i = 1; i <= n; i ++) {
            for(int j = 1; j <= n; j ++) {
                if(vis[i][k] == true && vis[k][j] == true)
                    vis[i][j] = true;
            }
        }
    }
    int ans = 0;
    for(int i = 1; i <= n; i ++) {
        bool flag = true;
        for(int j = 1; j <= n; j ++) {
            if(i == j)  continue;
            else if(!vis[i][j] && !vis[j][i]) {
                flag = false;
                break;
            }
        }
        if(flag == true)    ans ++;
    }
    printf("%d\n",ans);
    return 0;
}