POJ3275:Ranking the Cows(Bitset加速floyd求闭包传递)

Posted 2020-10-26 ---学习ing---

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 cowY.

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?"

 

题意：有N头奶牛，现在给出M对产奶量关系U>V，问至少还需要知道多少奶牛可以做到全部奶牛产奶关系。

思路：有向图，问至少再加多少边，使得任意两点S、T的可以到达（S到达T或者到达S）。闭包传递后不能到达的需要加边，ans++。

 

至于为什么闭包传递后不连通就就要ans++呢，假设已知了1>2>3>4，5>6>7，我们知道了4>5不就行了吗，ans=1啊。

但是注意题目说的是"确定"，而比较了4和5之后可能会4<5，即任然不可以把知道全部顺序。

所以剩下对于C(n,2)对关系里不确定的关系，都有知道才能“确定”所有的大小关系。

#include<cstdio>
#include<bitset>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn=1010;
bitset<maxn>mp[maxn];
int main()
{
    int N,M,ans,i,j,k;
    while(~scanf("%d%d",&N,&M)){
        ans=0;memset(mp,0,sizeof(mp));
        for(i=1;i<=M;i++){
            scanf("%d%d",&j,&k);
            mp[j].set(k);
        }
        
        for(k=1;k<=N;k++)
         for(i=1;i<=N;i++)
          if(mp[i][k])
           mp[i]|=mp[k];
        for(i=1;i<=N;i++)
         for(j=i+1;j<=N;j++)
          if(!mp[i][j]&&!mp[j][i])
           ans++;
        printf("%d\n",ans);
    }
    return 0;
}