学步园

给定二维平面内若干点的的坐标，问最近点对的距离。

暴力pass，必须是分治。可以把平面上的点分为左右两部分，最近的点对一定是左边的最近点对，右边的最近点对和左右分界线上的最近点对中距离最短的那一对。

题目对程序的速度性要求很高，跑t了好多发。

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#define SIZE 100000

using namespace std;

struct Node
{
        double x, y;
}p[SIZE];

int n;
int index[SIZE];

bool cmpx(const Node a, const Node b)
{
                return a.x < b.x;
}

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

double dis(const Node& a, const Node& b)
{
        double tmp = (a.x-b.x)*(a.x-b.x) + (a.y-b.y)*(a.y-b.y);
        return tmp;
}

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

double mindis(int left, int right)
{
        if (left + 1 == right)
                return dis(p[left], p[right]);
        else if (left + 2 == right)
                return m(dis(p[left], p[left+1]), m(dis(p[left+1], p[right]), dis(p[left], p[right])));
        else
        {
                int mid = (left + right) >> 1;
                double tmpdis = m(mindis(left, mid), mindis(mid+1, right));
                double a = p[mid].x - tmpdis;
                double b = p[mid].x + tmpdis;
                int len = 0;
                for (int i = left; i <= right; i++)
                {
                        if (p[i].x > a && p[i].x < b) index[len++] = i;
                }
                sort(index, index+len, cmpy);
                for(int i = 0; i < len; i++)
                {
                        for(int j = i+1; j < len; j++)
                        {
                                if (p[index[j]].y - p[index[i]].y > tmpdis) break;
                                tmpdis = m(tmpdis, dis(p[index[i]], p[index[j]]));
                        }
                }
                return tmpdis;
        }
}

int main()
{
        #ifdef BellWind
                freopen("1007.in", "r", stdin);
        #endif // BellWind

//        int cnt = 0;

        while (~scanf("%d", &n) && n)
        {
                for (int i = 0; i < n; i++) { scanf("%lf%lf", &p[i].x, &p[i].y); }
                sort(p, p+n, cmpx);
                double dis = mindis(0, n-1);
                printf("%.2lf\n", sqrt(dis)/2.0);
        }

        return 0;
}