(hdu1007)Quoit Design，求最近点对

Problem Description
Have you ever played quoit in a playground? Quoit is a game in which flat rings are pitched at some toys, with all the toys encircled awarded.
In the field of Cyberground, the position of each toy is fixed, and the ring is carefully designed so it can only encircle one toy at a time. On the other hand, to make the game look more attractive, the ring is designed to have the largest radius. Given a configuration of the field, you are supposed to find the radius of such a ring.

Assume that all the toys are points on a plane. A point is encircled by the ring if the distance between the point and the center of the ring is strictly less than the radius of the ring. If two toys are placed at the same point, the radius of the ring is considered to be 0.

Input
The input consists of several test cases. For each case, the first line contains an integer N (2 <= N <= 100,000), the total number of toys in the field. Then N lines follow, each contains a pair of (x, y) which are the coordinates of a toy. The input is terminated by N = 0.

Output
For each test case, print in one line the radius of the ring required by the Cyberground manager, accurate up to 2 decimal places.

Sample Input
2
0 0
1 1
2
1 1
1 1
3
-1.5 0
0 0
0 1.5
0

Sample Output
0.71
0.00
0.75

#include<map>#include<set>#include<queue>#include<stack>#include<vector>#include<math.h>#include<cstdio>#include<sstream>#include<numeric>//STL数值算法头文件#include<stdlib.h>#include <ctype.h>#include<string.h>#include<iostream>#include<algorithm>#include<functional>//模板类头文件using namespace std;typedef long long ll;const int maxn=10005;//const int INF=0x3f3f3f3f;const double INF = 1e20;const int N = 100005;int n;int tmpt[N];struct Point{    double x;    double y;} point[N];bool cmp(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);//分治求左边的最近点对    double d2 = Closest_Pair(mid+1,right);//分治求右边的最近点对    d = min(d1,d2);//求最小值    int i,j,k=0;    //分离出宽度为d的区间,求(m-d,m]和(m,m+d]之间的最近点对    //将在(m-d,m]范围内s1中的p和(m,m+d]范围内的点投影到一条直线    //然后将这些点按y坐标排序,进行线性扫描    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(~scanf("%d",&n)&&n)    {        for(int i = 0; i < n; i++)            scanf("%lf %lf",&point[i].x,&point[i].y);        sort(point,point+n,cmp);        printf("%.2lf\n",Closest_Pair(0,n-1)/2);    }    return 0;}