Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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

AC代码：

<span style="font-size:18px;">/*分治算法求最小点对*/
/*AC代码：828ms*/
#include <iostream>
#include <cmath>
#include <algorithm>
#define MAXN 100005
//#define min(a,b) (a<b?a:b)//为什么用这个就超时！！！
using namespace std;
struct Point
{
	double x,y;
};
struct Point point[MAXN],*px[MAXN],*py[MAXN];
double get_dis(Point *p1,Point *p2)
{
	return sqrt((p1->x-p2->x)*(p1->x-p2->x)+(p1->y-p2->y)*(p1->y-p2->y));
}
bool cmpx(Point *p1,Point *p2) {return p1->x<p2->x;}
bool cmpy(Point *p1,Point *p2) {return p1->y<p2->y;}
double min(double a,double b){return a<b?a:b;}
//-------核心代码------------//
double closest(int s,int e)
{
	if(s+1==e)
		return get_dis(px[s],px[e]);
	if(s+2==e)
		return min(get_dis(px[s],px[s+1]),min(get_dis(px[s+1],px[e]),get_dis(px[s],px[e])));
	int mid=(s+e)>>1;
	double ans=min(closest(s,mid),closest(mid+1,e));//递归求解
	int i,j,cnt=0;
	for(i=s;i<=e;i++)//把x坐标在px[mid].x-ans~px[mid].x+ans范围内的点取出来
	{
		if(px[i]->x >= px[mid]->x-ans && px[i]->x <= px[mid]->x+ans)
			py[cnt++]=px[i];
	}
	sort(py,py+cnt,cmpy);//按y坐标排序
	for(i=0;i<cnt;i++)
	{
		for(j=i+1;j<cnt;j++)//py数组中的点是按照y坐标升序的
		{
			if(py[j]->y-py[i]->y >= ans)
				break;
			ans=min(ans,get_dis(py[i],py[j]));
		}
	}
	return ans;
}
int main()
{
	int i,n;
	while(scanf("%d",&n)!=EOF)
	{
		if(n==0)
			break;
		for(i=0;i<n;i++)
		{
			scanf("%lf%lf",&point[i].x,&point[i].y);
			px[i]=&point[i];
		}
		sort(px,px+n,cmpx);
		//for(i=0;i<n;i++)
		// printf("(%.2lf,%.2lf)--",px[i].x,px[i].y);
		double distance=closest(0,n-1);
		printf("%.2lf\n",distance/2);
	}
	return 0;
}

</span>