HDU-1007-Quoit Design

#include<bits/stdc++.h>using namespace std;const double INF = 0x3f3f3f3f;struct Point{double x,y;}point[100005];int p[100005];bool cmpx(struct Point a,struct Point b){ return a.x < b.x; }bool cmpy(int a,int b){return point[a].y < point[b].y; } double get_distance(double x1,double y1,double x2,double y2){ return sqrt( pow(x1-x2,2) + pow(y1-y2,2) ); }double ClosetPoints(int left,int right)//S为平面上N个点的集合 {if(left == right) return INF;//只有一个点 if(left + 1 == right) return get_distance(point[left].x,point[left].y,point[right].x,point[right].y);//只有两个点 int mid = (left+right)>>1;double left_min = ClosetPoints(left,mid),right_min = ClosetPoints(mid + 1,right);//递归求解 double d = min(left_min,right_min);int k = 0;for(int i = left; i <= right; i++)//分离出x-d到x+d的区间 {if(fabs(point[i].x-point[mid].x)<=d){p[k++]=i;}}sort(p,p+k,cmpy);for(int i=0;i<k;i++){for(int j = i + 1; j < k && point[p[j]].y - point[p[i]].y < d; j++){d=min(d,get_distance(point[p[i]].x,point[p[i]].y,point[p[j]].x,point[p[j]].y));}}return d;} int main(){int n;while(~scanf("%d",&n)&&n){for(int i = 0; i < n; i++){scanf("%lf%lf",&point[i].x,&point[i].y);}sort(point,point+n,cmpx);printf("%.2lf\n",ClosetPoints(0,n-1)/2.0); }return 0;}/*int ClosestPoints(S){1.if(n<2) return ∞;2.m=s中各点x坐标的中位数(平均分为两个集合)3.  构造S1和S2,使得S1中点的x坐标小于m，S2中点的x坐标大于m；4.  d1=ClosestPoints(S1),d2=ClosestPoints(S);5.  d=min(d1,d2);6.  构造P1,P2,使得P1是S1中点的x坐标与m的距离小于d的点集,P2是S2中点的x坐标与m的距离小于d的点集7.  将P1,P2中的点按y坐标升序排序// 进行预排序会减少时间 8.  对P1中的每一个点p,在P2中查找与点p的y坐标小于d的点(最多6个点),并求出其中最小距离d39.  return min(d,d3); }*/