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
Source
#include <cstdio> #include <iostream> #include <string> #include <cstring> #include <cmath> #include <algorithm> #include <queue> #include <vector> #include <map> using namespace std; #define ll long long const int inf = 0x3f3f3f3f; int n, m, dis[100+8][100+8]; int main() { int a, b; memset(dis, 0, sizeof(dis)); scanf("%d%d", &n, &m); for(int i = 0; i<m; i++) { scanf("%d%d", &a, &b); dis[a][b] = 1; dis[b][a] = -1; } for(int i = 1; i <= n; i++) for(int j = 1; j <= n; j++) for(int k = 1; k <= n; k++) if(dis[j][i] == dis[i][k] && dis[j][i]) { dis[j][k] = dis[i][k]; } int ans = 0, sum; for(int i = 1; i <= n; i++) { sum = 0; for(int j = 1; j <= n; j++) if(dis[i][j] != 0) sum++; if(sum == n-1)ans++; } printf("%d\n", ans); return 0; }