HDOJ 1007 Quoit Design（分治）

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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 100000using 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;}

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