Cow Contest
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用warshall算法求传递闭包,最后求每个节点是否与剩下的n-1个节点是否有明确的关系并统计个数
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.
* 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
* Line 1: A single integer representing the number of cows whose ranks can be determined
5 54 34 23 21 22 5
2
#include<cstdio>#include<cstring>#define maxn 105using namespace std;bool an[maxn][maxn];int main(){int n,m;while(scanf("%d%d",&n,&m)!=EOF){memset(an,0,sizeof(an));for(int i=0;i<m;i++){int x,y;scanf("%d%d",&x,&y);an[y][x]=1;}for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){an[i][j]=an[i][j]|(an[i][k]&an[k][j]);}}}int j,ans=0;for(int i=1;i<=n;i++){for(j=1;j<=n;j++){if(i==j)continue;if(an[i][j]==0&&an[j][i]==0)break;}if(j>n)ans++;}printf("%d\n",ans);}return 0;}
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