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
根据每头牛之间的关系生成一张图, Warshall算法得到传递闭包。 若一头牛与其它每头牛之间关系确定, 则这头牛的排名就确定。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#define N 110
#define INF 0x3f3f3f3f
using namespace std;
bool Map[N][N];
int n, m;
int main()
{
while(cin >> n >> m)
{
memset(Map, false, sizeof(Map));
for(int i = 0; i < m; i++)
{
int u, v;
scanf("%d%d", &u, &v);
Map[u][v] = true;
}
for(int i = 1; i <= n; i++)// Warshall
{
for(int j = 1; j <= n; j++)
{
if(Map[j][i])
{
for(int k = 1; k <= n; k++)
{
Map[j][k] += Map[i][k];
}
}
}
}
int j;
int cnt = 0;
for(int i = 1; i <= n; i++)//枚举每头牛与其它牛间的关系, 只要与其中一只不能确定, 则不能确定其排名
{
for(j = 1; j <= n; j++)
{
if(i == j)
continue;
if(Map[i][j] || Map[j][i]);
else
break;
}
if(j > n)
cnt++;
}
printf("%d\n", cnt);
}
return 0;
}