POJ 3660 Cow Contest
Cow Contest
64-bit integer IO format: %lld Java class name: Main
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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 #include <climits> 7 #include <vector> 8 #include <queue> 9 #include <cstdlib> 10 #include <string> 11 #include <set> 12 #include <stack> 13 #define LL long long 14 #define pii pair<int,int> 15 #define INF 0x3f3f3f3f 16 using namespace std; 17 const int maxn = 101; 18 bool g[maxn][maxn]; 19 int n,m; 20 void Floyd() { 21 for(int k = 1; k <= n; ++k) { 22 for(int i = 1; i <= n; ++i) { 23 if(g[i][k]) { 24 for(int j = 1; j <= n; ++j) 25 if(!g[i][j]) g[i][j] = g[i][k]&&g[k][j]; 26 } 27 } 28 } 29 } 30 int main() { 31 while(~scanf("%d %d",&n,&m)) { 32 memset(g,false,sizeof(g)); 33 for(int i = 0; i < m; ++i) { 34 int u,v; 35 scanf("%d %d",&u,&v); 36 g[u][v] = true; 37 } 38 Floyd(); 39 int ans = 0,x,y; 40 for(int i = 1; i <= n; ++i){ 41 x = y = 0; 42 for(int j = 1; j <= n; ++j){ 43 if(g[i][j]) x++; 44 if(g[j][i]) y++; 45 } 46 if(x+y == n-1) ans++; 47 } 48 printf("%d\n",ans); 49 } 50 return 0; 51 }