POJ-3660.Cow Contest(有向图的传递闭包)

 

Cow Contest

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17797		Accepted: 9893

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

本题思路：将题目给出的已知边都存入图中，利用传递性求出可能存在的每条边，对于一个学生，用i -> j表示 i 比 j 强，那么对于所有学生，他的排名被确定的条件就是确定他与其它所有同学的排名情况，即Bigger[ i ] + Smaller[ i ] == n - 1。

参考代码：(不建议看，按照上面的思路实现一波就行了)

#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

const int maxn = 100 + 2, INF = 0x3f3f3f3f;
int n, m, a, b;
bool G[maxn][maxn];

int Floyd_Warshall() {
    int num = 0;
    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 ++) {
        int temp = 0;
        for(int j = 1; j <= n; j ++)
            if(G[i][j] || G[j][i]) temp ++;
        if(temp == n - 1) num ++;
    }
    return num;
}

int main () {
    scanf("%d %d", &n, &m);
    int x, y;
    for(int i = 0; i < m; i ++) {
        scanf("%d %d", &x, &y);
        G[x][y] = true;
    }
    int ans = Floyd_Warshall();
    printf("%d
", ans);
    return 0;
}