【分治 求最近点对】hdu 1007 Quoit Design

Link：http://acm.split.hdu.edu.cn/showproblem.php?pid=1007

#include <bits/stdc++.h>using namespace std;/*hdu 1007题意：给出物品在平面上的点坐标，求一个环不能一次套到两的最大半径，即最近点对距离的一半。题解：先以x排序，用分治将问题分成左边部分的最近点对，和左边的最近点对，左边右边各一个点的最近点对（在算这个的时候应按y排序，将两点y差大于当前最小值的优化掉）。*/const int N = 1e5+5;const double INF = 1e18;struct Point{double x,y; } a[N];bool cmp(Point a,Point b){return a.x < b.x; }bool cmp2(Point a,Point b){return a.y < b.y; }double dist(Point a,Point b){return sqrt((a.y-b.y)*(a.y-b.y)+(a.x-b.x)*(a.x-b.x))/2; }Point temp[N];double solve(int l, int r){    if(l == r)        return INF;    int mid = (l+r)>>1;    double mi;    mi = solve(l,mid);    mi = min(mi,solve(mid+1,r));//    printf("L=%f R=%f\n",solve(l,mid),solve(mid+1,r));    int top = 0;    temp[top++] = a[mid];    for(int i = mid-1; i >= l; i--){        if(a[mid].x-a[i].x>=mi)            break;        temp[top++] = a[i];    }    for(int i = mid+1; i <= r; i++){        if(a[i].x-a[mid].x>=mi)            break;        temp[top++] = a[i];    }    sort(temp,temp+top,cmp2);    for(int i = 0; i < top; i++){        for(int j = i+1; j < top; j++)        {            if(temp[j].y-temp[i].y>=mi)                break;            mi = min(mi,dist(temp[i],temp[j]));        }    }    return mi;}int main(){    int n;    while(~scanf("%d",&n) && n)    {        for(int i = 0; i < n; i++)            scanf("%lf%lf",&a[i].x,&a[i].y);        sort(a,a+n,cmp);        printf("%.2f\n",solve(0,n-1));    }    return 0;}