Cow Contest
时间限制:1000 ms | 内存限制:65535 KB
难度:4
- 描述
-
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.
- 输入
- * 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
There are multi test cases.The input is terminated by two zeros.The number of test cases is no more than 20. - 输出
- For every case:
* Line 1: A single integer representing the number of cows whose ranks can be determined - 样例输入
-
5 5 4 3 4 2 3 2 1 2 2 5 0 0
- 样例输出
-
2
////能打败的个数加上被打败的个数恰好等于n-1,则能确定, ////否则无法确定,抽象为简单的floyd传递闭包算法, ////在加上每个顶点的出度与入度 (出度+入度=顶点数-1,则能够确定其编号)。 #include<queue> #include<stack> #include<vector> #include<math.h> #include<stdio.h> #include<numeric>//STL数值算法头文件 #include<stdlib.h> #include<string.h> #include<iostream> #include<algorithm> #include<functional>//模板类头文件 using namespace std; const int INF=1e9+7; const int maxn=102; int n,m; int tu[maxn][maxn]; int main() { int i,j,k,x,y,ans; while(~scanf("%d %d",&n,&m),n+m) { memset(tu,0,sizeof(tu)); while(m--) { scanf("%d %d",&x,&y); tu[x][y]=1; } for(k=1; k<=n; k++) for(i=1; i<=n; i++) for(j=1; j<=n; j++) if(tu[i][k]&&tu[k][j]) tu[i][j]=1; int ans=0; for(i=1; i<=n; i++) { int cont=n-1; for(j=1; j<=n; j++) if(tu[i][j]||tu[j][i]) cont--; if(!cont) ans++; } printf("%d ",ans); } return 0; }