hdu 3585 二分+最大团

题目：给出平面上n个点，现在找m个点，并且使得这m个点最近的两个最远。

分析：显然这满足二分的性质，二分答案，根据点距离需要大于等于二分值重新构造新图，则问题变成了：在新图中找出满足所有点对之间的距离大于等于二分的值的一个子图。因此在新图中寻找最大团即可。具体看代码

#include <set>
#include <map>
#include <list>
#include <cmath>
#include <queue>
#include <stack>
#include <string>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;
typedef unsigned long long ull;

#define debug puts("here")
#define rep(i,n) for(int i=0;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define REP(i,a,b) for(int i=a;i<=b;i++)
#define foreach(i,vec) for(unsigned i=0;i<vec.size();i++)
#define pb push_back
#define RD(n) scanf("%d",&n)
#define RD2(x,y) scanf("%d%d",&x,&y)
#define RD3(x,y,z) scanf("%d%d%d",&x,&y,&z)
#define RD4(x,y,z,w) scanf("%d%d%d%d",&x,&y,&z,&w)
#define All(vec) vec.begin(),vec.end()
#define MP make_pair
#define PII pair<int,int>
#define PQ priority_queue
#define cmax(x,y) x = max(x,y)
#define cmin(x,y) x = min(x,y)
#define Clear(x) memset(x,0,sizeof(x))
/*

#pragma comment(linker, "/STACK:1024000000,1024000000")

int size = 256 << 20; // 256MB
char *p = (char*)malloc(size) + size;
__asm__("movl %0, %%esp\n" :: "r"(p) );

*/

char IN;
bool NEG;
inline void Int(int &x){
    NEG = 0;
    while(!isdigit(IN=getchar()))
        if(IN=='-')NEG = 1;
    x = IN-'0';
    while(isdigit(IN=getchar()))
        x = x*10+IN-'0';
    if(NEG)x = -x;
}
inline void LL(ll &x){
    NEG = 0;
    while(!isdigit(IN=getchar()))
        if(IN=='-')NEG = 1;
    x = IN-'0';
    while(isdigit(IN=getchar()))
        x = x*10+IN-'0';
    if(NEG)x = -x;
}

/******** program ********************/

const int MAXN = 50;

int g[MAXN][MAXN],n;
bool use[MAXN];
int dp[MAXN],best,m;

int px[MAXN],py[MAXN];
double dis[MAXN][MAXN];
//int pre[MAXN],path[MAXN]; // 记录路径

bool dfs(int *id,int top,int cnt){
    if(!top){
        if(best<cnt){
            //copy( pre+1,pre+cnt+1,path ); // 记录路径
            best = cnt;
            return true;
        }
        return false;
    }
    int a[MAXN];
    rep(i,top){
        if(cnt+top-i<=best)return false;
        if(cnt+dp[id[i]]<=best)return false;
        //pre[cnt] = id[i]; // 记录路径
        int k = 0;
        for(int j=i+1;j<top;j++)
            if(g[id[i]][id[j]])
                a[k++] = id[j];
        if(dfs(a,k,cnt+1))return true;
    }
    return false;
}

inline int solve(){
    int id[MAXN];
    best = 0;
    for(int i=n-1;i>=0;i--){
        int top = 0;
        for(int j=i+1;j<n;j++)
            if(g[i][j])
                id[top++] = j;
        dfs(id,top,1);
        dp[i] = best;
    }
    return best;
}

inline bool build(double mid){
    Clear(g);
    rep(i,n)
        rep(j,n)
            g[i][j] = dis[i][j]>=mid;
    return solve()>=m;
}

double cal(int x,int y){
    return sqrt(x*x+y*y+0.0);
}

int main(){

#ifndef ONLINE_JUDGE
    freopen("sum.in","r",stdin);
    //freopen("sum.out","w",stdout);
#endif

    while(~RD2(n,m)){
        rep(i,n)
            RD2(px[i],py[i]);
        rep(i,n)
            for(int j=i+1;j<n;j++)
                dis[i][j] = dis[j][i] = cal(px[i]-px[j],py[i]-py[j]);
        double l = 0 , r = 10000000;
        double ans = 0;
        rep(step,50){
            double mid = (l+r)/2;
            if(build(mid)){
                l = mid;
                ans = mid;
            }else r = mid;
        }
        printf("%.2lf\n",ans);
    }

    return 0;
}