HDU3585 最大团+二分

maximum shortest distance

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1777    Accepted Submission(s): 589

Problem Description

There are n points in the plane. Your task is to pick k points (k>=2), and make the closest points in these k points as far as possible.

 

Input

For each case, the first line contains two integers n and k. The following n lines represent n points. Each contains two integers x and y. 2<=n<=50, 2<=k<=n, 0<=x,y<10000.

 

Output

For each case, output a line contains a real number with precision up to two decimal places.

 

Sample Input

3 2

0 0

10 0

0 20

 

Sample Output

22.36

 

Author

alpc50

题意：

给出n个点，问从中取哪k个点使得这k个点中每两个点之间的距离的最小值最大，输出这个距离。

代码：

//求最小值最大一般用二分。此题求出每两点之间的距离，然后二分距离，两点之间的距离大于等于此二分值的点之间连线，
//重新建图，求新图的最大团是否大于等于k。这道题可以不直接二分距离，给距离排序二分下标即可。套最大团模板。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
int n,best,m;
int mp[][],num[];
double dis[],tmp[][],x[],y[];
bool dfs(int s[],int sum,int cnt)
{
    if(sum==){
        if(cnt>best){
            best=cnt;
            return ;
        }
        return ;
    }
    int t[];
    for(int i=;i<sum;i++){
        if(cnt+(sum-i)<=best) return ;
        if(cnt+num[s[i]]<=best) return ;
        int k=;
        for(int j=i+;j<sum;j++){
            if(mp[s[i]][s[j]]) t[k++]=s[j];
        }
        if(dfs(t,k,cnt+)) return ;
    }
    return ;
}
bool Maxt(int M)
{
    //if(n<=0) return 0;
    int s[];
    best=;
    memset(mp,,sizeof(mp));
    for(int i=;i<n;i++)
        for(int j=i+;j<n;j++)
            if(tmp[i][j]>=dis[M]) mp[i][j]=mp[j][i]=;
    for(int i=n-;i>=;i--){
        int k=;
        for(int j=i+;j<n;j++){
            if(mp[i][j]) s[k++]=j;
        }
        dfs(s,k,);
        num[i]=best;
        if(best>=m) return true;
    }
    return false;
}
int main()
{
    while(~scanf("%d%d",&n,&m)){
        int cnt=;
        for(int i=;i<n;i++){
            scanf("%lf%lf",&x[i],&y[i]);
            tmp[i][i]=;
            for(int j=;j<i;j++)
                dis[cnt++]=tmp[i][j]=tmp[j][i]=((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
        }
        sort(dis,dis+cnt);
        int ans=;
        int L=,R=cnt-,M;
        while(L<=R){
            M=(L+R)/;
            if(Maxt(M)) {L=M+;ans=M;}
            else R=M-;
        }
        printf("%.2lf\n",sqrt(dis[ans]));
    }
    return ;
}