poj 3660 Cow Contest

Cow Contest

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7016 Accepted: 3853

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 cowA 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

思路：先判断是否联通 然后分别统计个点的入度和出度之和靠是否为n-1 即可 传递闭包的性质出度和入度和为n-1的时候其顺序是唯一确定的

#include <iostream>#include <cstdio>#include <cstring>#include <cmath>#include <algorithm>#include <stack>#include <queue>#include <set>#include <vector>using namespace std;#define maxn  105#define INF  1000000000int d[maxn][maxn];int n,m; void floyd(){int i,j,k;for(i=1;i<=n;i++){for(j=1;j<=n;j++){for(k=1;k<=n;k++){//if(d[j][i]!=-1&&d[i][k]!=-1)//if(d[j][k]==-1)d[j][k]=d[j][k]||(d[j][i]&&d[i][k]);}}}}int main(){int i,j;int a,b;scanf("%d %d",&n,&m);    memset(d,0,sizeof(d));    for(i=0;i<m;i++){      scanf("%d %d",&a,&b);       d[a][b]=1;}floyd();int cnt=0,k=0;for(i=1;i<=n;i++){  k=0;for(j=1;j<=n;j++){  if(d[i][j]||d[j][i])  k++;  }if(k==n-1)cnt++;}cout<<cnt<<endl;return 0;}