hdu3585 maximum shortest distance

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maximum shortest distance

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 20 010 00 20

Sample Output

22.36

Author

alpc50

Source

2010 ACM-ICPC Multi-University Training Contest（15）——Host by NUDT

题意:在N个点钟选出K个点使得这K个点间的最小距离最大

思路:二分答案+最大团

#include<bits/stdc++.h>using namespace std;#define Hash(a,b) (a-1)*n+bconst int maxn=51;double mp[maxn][maxn];int num[maxn],k;int G[maxn][maxn],ret;bool dfs(int *adj,int tot,int cnt){    if(cnt>=k){        ret=cnt;        return 1;    }    if(tot==0){        if(cnt>ret){            ret=cnt;            return 1;        }        return 0;    }    for(int i=0;i<tot;i++){        if(cnt+tot-i<=ret)            continue;        if(cnt+num[adj[i]]<=ret)            continue;        int t[maxn],k=0;        for(int j=i+1;j<tot;j++){            if(G[adj[i]][adj[j]])                t[k ++] = adj[j];        }        if(dfs(t,k,cnt+1))            return 1;    }    return 0;}void Maxclique(int n){    int adj[maxn],tot;    for(int i=n;i>=1;i--){        tot=0;        for(int j=i+1;j<=n;j++)            if(G[i][j]==1)                adj[tot++]=j;        dfs(adj,tot,1);        if(ret>=k)            return ;        num[i]=ret;    }}struct Point{    double x,y;}node[maxn];double distance(int i,int j){    return sqrt((node[i].y-node[j].y)*(node[i].y-node[j].y)+(node[i].x-node[j].x)*(node[i].x-node[j].x));}int main(){    int _;    int n;    while(scanf("%d%d",&n,&k)!=EOF){        memset(G,0,sizeof(G));        ret=0;        for(int i=1;i<=n;i++)            scanf("%lf%lf",&node[i].x,&node[i].y);        for(int i=1;i<=n;i++)            for(int j=1;j<=n;j++)                mp[i][j]=distance(i,j);        double low=0,high=20000;        while(high-low>=1e-4){            ret=0;            double mid=(low+high)/2;            for(int i=1;i<=n;i++)                for(int j=1;j<=n;j++){                    if(mp[i][j]>=mid)                        G[i][j]=1;                    else                        G[i][j]=0;                }            Maxclique(n);            if(ret>=k)                low=mid+(1e-4);            else                high=mid-(1e-4);        }        printf("%.2f\n",(high+low)/2);    }    return 0;}

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