poj 3660

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 10890 Accepted: 6065

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

题解：如果1可以打败2,2可以打败3，那么1可以打败3，所以只要确定某一个数与其他n-1个之间的关系，可能是打败，也可能是被打败，就能确定他的名字。

#include<stdio.h>int map[105][105];void floya(int n) {int i, j, k;for(k = 1; k <= n; k++)for(i = 1; i <= n; i++)for(j = 1; j <= n; j++)if(map[i][k] && map[k][j])map[i][j] = 1;}int main() {int n, m;while (scanf("%d%d", &n, &m) != EOF) {int i, j;for(i = 1; i < 105; i++)for(j = 1; j < 105; j++)map[i][j] = 0;int count, win, down, sum = 0;for(i = 1; i <= m; i++) {scanf("%d%d", &win, &down);map[win][down] = 1;}floya(n);for(i = 1; i <= n; i++) {count = 0;for(j = 1; j <= n; j++) if(map[i][j] || map[j][i])count++;if(count == n - 1)sum++;}printf("%d\n", sum);}}