POJ 3660 Cow Contest(传递闭包)
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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 54 34 23 21 22 5
Sample Output
2
Source
题目大意:
给出N个人的合法的强弱关系,问最多能确定多少人的排名。
解题思路:
对于任意一个人,如果比他强的人与比他弱的人之和为N-1,那么就可以确定他的排名。
对于有多少人比他强或比他弱,我们可以用floyd求传递闭包来得到。
AC代码:
#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <vector>#include <queue>#include <map>#include <cmath>using namespace std;#define INF 0x3f3f3f3f#define fi first#define se second#define mem(a,b) memset((a),(b),sizeof(a))const int MAXV=100+3;int V,E;//顶点数,边数int cnt[MAXV];//比他强的人和比他弱的人之和bool G[MAXV][MAXV];//图void init()//初始化{ for(int u=1;u<=V;++u) { for(int v=1;v<=V;++v) G[u][v]=u==v; cnt[u]=0; }}int main(){ while(~scanf("%d%d",&V,&E)) { init(); for(int i=0;i<E;++i) { int u,v; scanf("%d%d",&u,&v); G[u][v]=true; } for(int k=1;k<=V;++k)//floyd for(int u=1;u<=V;++u) for(int v=1;v<=V;++v) G[u][v]|=G[u][k]&G[k][v]; for(int u=1;u<=V;++u) for(int v=1;v<=V;++v) if(u!=v&&G[u][v]) { ++cnt[u]; ++cnt[v]; } int ans=0; for(int i=1;i<=V;++i) ans+=cnt[i]==V-1; printf("%d\n",ans); } return 0;}
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