分治法求最近点对

/**
最近点对问题，时间复杂度为O(n*logn*logn)
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const double INF = 1e20;
const int N = 100005;

struct Point
{
    double x;
    double y;
}point[N];
int n;
int tmpt[N];

bool cmpxy(const Point& a, const Point& b)
{
    if(a.x != b.x)
        return a.x < b.x;
    return a.y < b.y;
}

bool cmpy(const int& a, const int& b)
{
    return point[a].y < point[b].y;
}

double min(double a, double b)
{
    return a < b ? a : b;
}

double dis(int i, int j)
{
    return sqrt((point[i].x-point[j].x)*(point[i].x-point[j].x)
                + (point[i].y-point[j].y)*(point[i].y-point[j].y));
}

double Closest_Pair(int left, int right)
{
    double d = INF;
    if(left==right)
        return d;
    if(left + 1 == right)
        return dis(left, right);
    int mid = (left+right)>>1;
    double d1 = Closest_Pair(left,mid);
    double d2 = Closest_Pair(mid+1,right);
    d = min(d1,d2);
    int i,j,k=0;
    //分离出宽度为d的区间
    for(i = left; i <= right; i++)
    {
        if(fabs(point[mid].x-point[i].x) <= d)
            tmpt[k++] = i;
    }
    sort(tmpt,tmpt+k,cmpy);
    //线性扫描
    for(i = 0; i < k; i++)
    {
        for(j = i+1; j < k && point[tmpt[j]].y-point[tmpt[i]].y<d; j++)
        {
            double d3 = dis(tmpt[i],tmpt[j]);
            if(d > d3)
                d = d3;
        }
    }
    return d;
}


int main()
{
    while(true)
    {
        scanf("%d",&n);
        if(n==0)
            break;
        for(int i = 0; i < n; i++)
            scanf("%lf %lf",&point[i].x,&point[i].y);
        sort(point,point+n,cmpxy);
        printf("%.2lf\n",Closest_Pair(0,n-1)/2);
    }
    return 0;
}