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

Author

CHEN, Yue

Source

ZJCPC2004

这道题正规的做法是分治，不过因为数据极弱的关系，我普通方法就过了。题意：给你N个点的坐标，叫你找出哪两个点距离最近，然后距离除以2就是所求。。

思路：按X升序排，Y升序排，比较两种的结果哪种距离更小。可得解，跑了一千多毫秒。

上代码解析。

#include <stdio.h>
#define min(a,b) a>b?b:a;
#include <math.h>
#include <algorithm>
using namespace std;
struct point
{
	double x,y;  //必须用double 因为必然有小数点
}
f[100005];
bool cmp(point a,point b)
{
	if(fabs(a.x-b.x)<1e-8)  // 如果两个坐标的距离很接近
		return a.y<b.y;  //  按Y的升序排
	return a.x<b.x;//   X也升序。
}
bool cmp1(point a,point b)
{

	if(fabs(a.y-b.y)<1e-8)// 同上
		return a.x<b.x;
	return a.y<b.y;
}
double find(double a,double b,double c,double d)
{
	return sqrt(pow((a-c),2)+pow((b-d),2));//  计算两点之间的距离
}
void solve()
{
    int n,i;
	double x,y;
	while(scanf("%d",&n)&&n)
	{
		for(i=0;i<n;i++)
			scanf("%lf%lf",&f[i].x,&f[i].y);
		sort(f,f+n,cmp);//  第一种排序
		double p=99999999;
		double s;
		for(i=0;i<n-1;i++)
		{
			s=min(p,find(f[i].x,f[i].y,f[i+1].x,f[i+1].y));//找出最小的距离
			p=s;
		}
		sort(f,f+n,cmp1);// 第二种排序
		   	  for(i=0;i<n-1;i++)
			  {
				  s=min(p,find(f[i].x,f[i].y,f[i+1].x,f[i+1].y));
				  p=s;//找出第一种和第二种哪个距离更小。
			  }
			  s/=2;//除以2就是答案
			  printf("%.2lf\n",s);输出
	}
}
int main()
{
	solve();//问题解决
	return 0;
}