| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5989 | Accepted: 3234 |
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：是给你n个人，m组关系
每次关系输入A,B。表示A比B屌
思路：Flyod跑一发，然后判断是否这个人和其他人的关系都已经确定，如果都已经确定，那么就直接ans++就好了！
//qscqesze #include <cstdio> #include <cmath> #include <cstring> #include <ctime> #include <iostream> #include <algorithm> #include <set> #include <vector> #include <sstream> #include <queue> #include <typeinfo> #include <fstream> #include <map> typedef long long ll; using namespace std; //freopen("D.in","r",stdin); //freopen("D.out","w",stdout); #define sspeed ios_base::sync_with_stdio(0);cin.tie(0) #define maxn 105 #define eps 1e-9 const int inf=0x7fffffff; //无限大 int g[maxn][maxn]; int main() { int n,m; while(cin>>n>>m) { int a,b; for(int i=0;i<m;i++) { cin>>a>>b; g[a][b]=1; } for(int k=1;k<=n;k++) { for(int i=1;i<=n;i++) { for(int j=1;j<=n;j++) { if(g[i][k]&&g[k][j]) g[i][j]=1; } } } int ans=0; for(int i=1;i<=n;i++) { int flag=1; for(int j=1;j<=n;j++) { if(i==j) continue; if(g[i][j]==0&&g[j][i]==0) { flag=0; break; } } if(flag) ans++; } cout<<ans<<endl; } }