Cow Contest——Floyed+连通性判断

【题目描述】
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.
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

【题目分析】
需要将题目转化成图的问题，假如我们我打赢看作一条有向边，那么对于每个节点，我们判断其他节点能否到达他（比他厉害）或者他能否到达其他节点（比他菜），如果所有节点都可以，那么他的排名就是唯一确定的，所以我们用一个Floyed判断一下

【AC代码】

#include<cstdio> #include<cstring> #include<cstdlib> #include<algorithm> #include<iostream> #include<cmath> #include<climits> #include<queue> #include<vector> #include<set> #include<map> using namespace std;typedef long long ll; const int MAXN=105; bool a[MAXN][MAXN],flag; int n,m,ans;void Floyed() {for(int k=1;k<=n;k++){for(int i=1;i<=n;i++){for(int j=1;j<=n;j++){a[i][j]=a[i][j] || (a[i][k] && a[k][j]);}}} }int main() {int u,v;while(~scanf("%d%d",&n,&m)){memset(a,0,sizeof(a));ans=0;for(int i=0;i<m;++i){scanf("%d%d",&u,&v);a[u][v]=true;}Floyed();for(int i=1;i<=n;i++){flag=true;for(int j=1;j<=n;j++){if(i==j) continue;if(!a[i][j] && !a[j][i]){flag=false;break;}}if(flag) ans++;}printf("%d\n",ans);}return 0; }

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