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Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47126    Accepted Submission(s): 12323

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

思路：分治；

最近点对模板题；

这个题目其实就是求最近点对的距离。《算法导论》上有详细讲解，王晓东的书上也有代码。主要思想就是分治。先把n个点按x坐标排序，然后求左边n/2个和右边n/2个的最近距离，最后合并。

合并要重点说一下，比较麻烦。

首先，假设点是n个，编号为1到n。我们要分治求，则找一个中间的编号m，先求出1到m点的最近距离设为d1，还有m+1到n的最近距离设为d2。这里的点需要按x坐标的顺序排好，并且假设这些点中，没有2点在同一个位置。（若有，则直接最小距离为0了）。

然后，令d为d1, d2中较小的那个点。如果说最近点对中的两点都在1-m集合中，或者m+1到n集合中，则d就是最小距离了。但是还有可能的是最近点对中的两点分属这两个集合，所以我们必须先检测一下这种情况是否会存在，若存在，则把这个最近点对的距离记录下来，去更新d。这样我们就可以得道最小的距离d了。

关键是要去检测最近点对，理论上每个点都要和对面集合的点匹配一次，那效率还是不能满足我们的要求。所以这里要优化。怎么优化呢？考虑一下，假如以我们所选的分割点m为界，如果某一点的横坐标到点m的横坐标的绝对值超过d1并且超过d2，那么这个点到m点的距离必然超过d1和d2中的小者，所以这个点到对方集合的任意点的距离必然不是所有点中最小的。（上面的是噢别人的，链接http://blog.csdn.net/allenjy123/article/details/6627751）

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<queue>
 5 #include<string.h>
 6 #include<stdlib.h>
 7 #include<set>
 8 #include<math.h>
 9 using namespace std;
10 const double INF=1e12;
11 typedef struct pp
12 {
13     double x;
14     double y;
15 } ss;
16 bool cmp_x(pp p,pp q)
17 {
18     return p.x < q.x;
19 }
20 bool cmp_y(pp p,pp q)
21 {
22     return p.y < q.y;
23 }
24 double slove(pp *s,int m);
25 ss node[1000005];
26 int main(void)
27 {
28     int i,j;
29     int n;int __ca=0;
30     while(scanf("%d",&n),n != 0)
31     {
32         for(i = 0; i < n; i++)
33         {
34             scanf("%lf %lf",&node[i].x,&node[i].y);
35         }
36         sort(node,node+n,cmp_x);
37         double ask = slove(node,n);
38         printf("%.2f\n",ask/2);
39     }
40     return 0;
41 }
42 double slove(ss *s,int m)
43 {
44     if(m <= 1)return INF;
45     int mid = m/2;
46     double d;int x = s[mid].x;
47     d = min(slove(s,mid),slove(s+mid,m-mid));
48     inplace_merge(s,s+mid,s+m,cmp_y);
49     int i,j;
50     vector<ss>vec;
51     for(i = 0;i < m; i++)
52     {
53         if(fabs(s[i].x-x) >= d)
54             continue;
55         int cn = vec.size();
56         for(j = 0;j < cn ;j++)
57         {
58             double dx = fabs(s[i].x - vec[cn-j-1].x);
59             double dy = fabs(s[i].y - vec[cn-j-1].y);
60             if(dy >= d)
61                 break;
62             d = min(d,sqrt(dx*dx+dy*dy));
63         }
64         vec.push_back(s[i]);
65     }
66     return d;
67 }