Floyd传递闭包-POJ-3660-Cow Contest

Cow Contest
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 8605 Accepted: 4834
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

求能够确定rank的牛牛的数量，那么就是求与其他n-1只牛都有明确关系的牛的数量，用floyd传递闭包解决。

////  main.cpp//  最短路练习-H-Cow Contest////  Created by 袁子涵 on 15/10/14.//  Copyright © 2015年 袁子涵. All rights reserved.//  32ms    780KB#include <iostream>#include <string.h>#include <stdlib.h>#include <stdio.h>#include <algorithm>#include <math.h>#define MAXN 105#define INF 0x3f3f3f3fusing namespace std;int N,M,sum=0;int dis[MAXN][MAXN];int main(int argc, const char * argv[]) {    cin >> N >> M;    memset(dis, 0, sizeof(dis));    int a,b;    for (int i=1; i<=M; i++) {        scanf("%d %d",&a,&b);        dis[a][b]=1;        dis[b][a]=-1;    }    for (int i=1; i<=N; i++) {        for (int j=1; j<=N; j++) {            for (int k=1; k<=N; k++) {                if (dis[j][k]==0 && abs(dis[j][i]+dis[i][k])==2) {                    dis[j][k]=dis[j][i];                }            }        }    }    for (int i=1; i<=N; i++) {        for (int j=1; j<=N; j++) {            if (dis[i][j]==0 && i!=j) {                break;            }            if (j==N) {                sum++;            }        }    }    cout << sum << endl;    return 0;}