| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14455 | Accepted: 8097 |
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
//题意:给你n头牛，和m种两头牛之间的关系; 然后接下来给你这m种关系,每种关系有a, b两个数构成,表示a牛可以击败b牛；现在问这n头牛有几头牛的排名能够确定;
//比如案例中的2号牛和5号牛，分别就是倒数第二，和倒数第一,这两个可以确定，所以输出2;
//理解:如果当前牛与其他所有牛都有关系（直接或者间接），那么这头牛的排名就可以确定，这是本题的核心（画案例中的图就知道）；构建一个关系图，通过floyd算法来完成这n个点之间的闭包关系;
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<vector>
#include<cmath>
#define maxn 105
using namespace std;
int G[maxn][maxn];
int n, m;
void floyd(){
for(int k = 1; k <= n; k++){
for(int i = 1; i <= n; i++){
for(int j = 1; j <= n; j++){ //看看任意两头牛的关系如何,如果有关系值就是1(包括之间关系和间接关系)
G[i][j] = G[i][j] || (G[i][k]&&G[k][j]); //前面G[i][j]表示i,j有直接关系,(G[i][k]&&G[k][j])表示i,j可以通过其他牛找到间接关系;
}
}
}
return;
}
int main(){
int a, b;
while(scanf("%d%d", &n, &m) != EOF){
for(int i = 0; i <= n; i++){
for(int j = 0; j <= n; j++)
G[i][j] = 0;
}
for(int i = 0; i < m; i++){
scanf("%d%d", &a, &b);
G[a][b] = 1;
}
for(int i = 1; i <= n; i++){ //自己和自己有关系,这一步不能丢了!
G[i][i] = 1;
}
floyd();
int flag = 1;
int ans = 0;
for(int i = 1; i <= n; i++){
flag = 1;
for(int j = 1; j <= n; j++){ //对于第i头牛，我们通过这个for循环看看它是不是和其他所有牛都有关系;
flag = flag && (G[i][j] || G[j][i]); //只要有一个关系不能确定，flag就是0,只有所有关系都确定是flag的值才是1;
}
if(flag)
ans++; //能够与所有牛确定关系的牛有几头;
}
cout << ans << endl;
}
return 0;
}