Cow Contest
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 16941 | Accepted: 9447 |
题目链接：http://poj.org/problem?id=3660
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：
有n个人，m场比赛，然后给出m场比赛的胜负关系，问有多少只牛能确定它们自己的名次。
题解：
这个题有点像拓扑排序，但是只用拓扑序并不能保证结果的正确性。
其实解这个题我们只需要发现这样一个关系就好了，若一只牛的名次能够被确定，那么它赢它的牛和它赢的牛个数之和为n-1。
利用这个关系，我们floyd传递闭包预处理一下，然后判断一下数量关系就好了。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <queue>
using namespace std;
typedef long long ll;
const int N = 105, M = 4505;
int n,m;
int mp[N][N];
int main(){
scanf("%d%d",&n,&m);
for(int i=1;i<=m;i++){
int u,v;
scanf("%d%d",&u,&v);
mp[u][v]=1;
}
for(int k=1;k<=n;k++){
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
mp[i][j]=(mp[i][j]|(mp[i][k]&mp[k][j]));
}
}
}
int ans=0;
for(int i=1;i<=n;i++){
int win=0,lose=0;
for(int j=1;j<=n;j++){
if(mp[i][j]) win++;
if(mp[j][i]) lose++;
}
if(win+lose==n-1) ans++;
}
cout<<ans;
return 0;
}
原文地址：https://www.cnblogs.com/heyuhhh/p/10352095.html