*HDU 1007 计算几何

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 48729    Accepted Submission(s): 12823

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

 

题意：

求点集中相距最小的两个点，他们之间的距离的一半。

代码：

 //分治法求最近点对模板。详解《算法导论》610页。//至于将y坐标只预排序一次怎么写呢。
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 int n;
 int a[];
 struct nod
 {
     double x,y;
 }p[];
 bool cmpx(nod p1,nod p2)
 {
     return p1.x<p2.x;
 }
 bool cmpy(int a1,int a2)
 {
     return p[a1].y<p[a2].y;
 }
 double dis(int s,int e)
 {
     return sqrt((p[s].x-p[e].x)*(p[s].x-p[e].x)+(p[s].y-p[e].y)*(p[s].y-p[e].y));
 }
 double Min(double a1,double a2,double a3)
 {
     return a1<a2?(a1<a3?a1:a3):(a2<a3?a2:a3);
 }
 double fenzhi(int s,int e)
 {
     if(s+==e)
     return dis(s,e);
     if(s+==e)
     return Min(dis(s,s+),dis(s,e),dis(s+,e));
     int mid=(s+e)>>;
     double ans=min(fenzhi(s,mid),fenzhi(mid+,e));
     int k=;
     for(int i=s;i<=e;i++)
     {
         if(fabs(p[i].x-p[mid].x)<ans)
         a[k++]=i;
     }
     sort(a,a+k,cmpy);
     for(int i=;i<k;i++)
     for(int j=i+;j<=i+&&j<k;j++)
     {
         if(p[a[j]].y-p[a[i]].y>=ans) break;
         else ans=min(ans,dis(a[i],a[j]));
     }
     return ans;
 }
 int main()
 {
     while(scanf("%d",&n)&&n)
     {
         for(int i=;i<n;i++)
         scanf("%lf%lf",&p[i].x,&p[i].y);
         sort(p,p+n,cmpx);
         double ans=fenzhi(,n-);
         printf("%.2lf\n",ans/);
     }
     return ;
 }