poj 3660 Cow Contest(最短路floyd)

题意：问我们能确定的牛的等级的个数，如果一个牛可以打败x个，被y个打败，当x+y=n-1时，就可以确定这个牛的等级了，因为要看所有牛之间能否联通或被联通，所有用floyd算法判断。

Cow ContestTime Limit:1000MS    Memory Limit:65536KB    64bit IO Format:%lld & %llu

SubmitStatusPracticePOJ 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 cowA has a greater skill level than cow B (1 ≤ A ≤ N; 1 ≤B ≤ N; A ≠ B), then cow A will always beat cowB.

Farmer John is trying to rank the cows by skill level. Given a list the results ofM (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

AC代码：

#include<iostream>#include<cstdio>#include<cmath>#include<algorithm>#include<cstring>#include<string>#include<vector>#include<queue>#include<map>#include<set>#define INF 100005#define LL long long#define mod 1000003using namespace std;int n, m;int d[105][105];int main(){while (scanf("%d%d", &n, &m) != EOF){int a, b;for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)d[i][j] = INF;for (int i = 0; i < m; i++){scanf("%d%d", &a, &b);d[a][b]= 1;}for (int k = 1; k <= n; k++)for (int i = 0; i <= n; i++)for (int j = 0; j <= n; j++)d[i][j] = min(d[i][j], d[i][k] + d[k][j]);int k=0;for (int i = 1; i <= n; i++){int f = 0;for (int j = 1; j <= n; j++){if (d[i][j] <INF||d[j][i]<INF){f++;}}if (f == n-1){k++;}}printf("%d\n", k);}}