POJ 3660 Cow Contest

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10114 Accepted: 5720

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 54 34 23 21 22 5

Sample Output

2

如果一个牛获胜数*加上失败数*等于总牛数减一，那么我们便可以确定他的排名了。获胜和失败包括间接获胜或失败例如：1号胜了2号，2号胜了5号，则5号输给了1号和2号。


利用递推式我们可以得出任意两个牛之间获胜或失败的关系。（这难道是用了任意两点间的最短路问题的思想！？什么，连函数名都一样。好吧，我承认函数名一样，不过 这不重要。这是技巧上的相似 而不是思想上的相似）

d[i][j] =d[i][j] || (d[i][k] & d[k][j])

#include <iostream>#include <cstdio>using namespace std;int n, m;int d[120][120];void floyd(){    for (int k = 1; k <= n; k++)        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                d[i][j] =d[i][j] || (d[i][k] & d[k][j]);}int main(){    while (cin >> n >> m)    {        int ans = 0, a, b;        for (int i = 1; i <= n; i++)            for (int j = 1; j <= n; j++)                d[i][j] = 0;        for (int i = 0; i < m; i++)        {            cin >> a >> b;            d[a][b] = 1;        }        floyd();        for (int i = 1; i <= n; i++)        {            int cnt = 0;            for (int j = 1; j <= n; j++)            {                if (d[i][j] || d[j][i])                    cnt++;            }            if (cnt == n-1)                ans++;        }        cout << ans << endl;    }    return 0;}