POJ3275 Ranking the Cows题解

题目

链接

http://poj.org/problem?id=3275

字面描述

Ranking the Cows
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 3893 Accepted: 1754
Description

Each of Farmer John’s N cows (1 ≤ N ≤ 1,000) produces milk at a different positive rate, and FJ would like to order his cows according to these rates from the fastest milk producer to the slowest.

FJ has already compared the milk output rate for M (1 ≤ M ≤ 10,000) pairs of cows. He wants to make a list of C additional pairs of cows such that, if he now compares those C pairs, he will definitely be able to deduce the correct ordering of all N cows. Please help him determine the minimum value of C for which such a list is possible.

Input

Line 1: Two space-separated integers: N and M
Lines 2…M+1: Two space-separated integers, respectively: X and Y. Both X and Y are in the range 1…N and describe a comparison where cow X was ranked higher than cow Y.
Output

Line 1: A single integer that is the minimum value of C.
Sample Input

5 5
2 1
1 5
2 3
1 4
3 4
Sample Output

3
Hint

From the information in the 5 test results, Farmer John knows that since cow 2 > cow 1 > cow 5 and cow 2 > cow 3 > cow 4, cow 2 has the highest rank. However, he needs to know whether cow 1 > cow 3 to determine the cow with the second highest rank. Also, he will need one more question to determine the ordering between cow 4 and cow 5. After that, he will need to know if cow 5 > cow 3 if cow 1 has higher rank than cow 3. He will have to ask three questions in order to be sure he has the rankings: “Is cow 1 > cow 3? Is cow 4 > cow 5? Is cow 5 > cow 3?”
Source

USACO 2007 March Gold

代码实现

#include<cstdio>
#include<bitset>
using namespace std;

const int maxn=1e3+10;
int n,m,ans;
bitset<maxn>mp[maxn];
int main(){
    
	scanf("%d%d",&n,&m);
	for(int i=1;i<=m;i++){
    
		int x,y;
		scanf("%d%d",&x,&y);
		mp[x][y]=1;
	}
	for(int j=1;j<=n;j++){
    
		for(int i=1;i<=n;i++){
    
			if(mp[i][j])mp[i]|=mp[j];
		}
	}
	for(int i=1;i<=n;i++){
    
		for(int j=1;j<=n;j++){
    
			if(mp[i][j])ans++;
		}
	}
	printf("%d\n",n*(n-1)/2-ans);
	return 0;
}