ACM-計算幾何学のQuoit Design-hdu 1007 zoj 2107

Quoit Design
Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 28539    Accepted Submission(s): 7469
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
 
テーマ:hdu 1007    ,       zoj  2107
この問題、タイプ:最近のポイントのペアを求めます.
平面の最近の点対を求める方法は、分治法です.
まず、点を2つの区間に分けて、S 1,S 2と仮定し、S 1内の最近接点対の点d 1,S 2内の最近接点対の点d 2をそれぞれ求める
さらにS 1とS 2内の最近の点対d=min(d 1,d 2)を求める
しかし、最近の点対はS 1点でS 2点である可能性があることを忘れてはいけない.
次はエッセンスを比較する部分です.
求めた点の位置は、きっと mid-d,mid+dの間.
そして、この区間でポイントを探し始め、d値を更新し続け、最後にdを得ることができます.
この問題は、最終的には半径が要求されるので、2で割る必要があります.

/**************************************
***************************************
*        Author:Tree                  *
*From :http://blog.csdn.net/lttree    *
* Title : Quoit Design                *
*Source: hdu 1007 zoj 2107            *
* Hint  :     ——             *
***************************************
**************************************/
#include <stdio.h>
#include <math.h>
#include <algorithm>
using namespace std;
#define N 100001
struct Point
{
    double x,y;
}p[N];
int arr[N];
double Min(double a,double b)
{
    return a<b?a:b;
}
//         
double dis(Point a,Point b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
//       or     
bool cmp_y( int a,int b)
{
    return p[a].y<p[b].y;
}
bool cmp_x( Point a,Point b)
{
    return a.x<b.x;
}
//      
double close_pair( int l,int r )
{
    //             
    if( r==l+1 )    return dis( p[l],p[r] );
    else if( r==l+2 )   return Min( dis(p[l],p[r]),Min( dis(p[l],p[l+1]),dis(p[l+1],p[r]) ) );

    int mid=(l+r)>>1;
    double ans=Min(close_pair(l,mid),close_pair(mid+1,r));

    int i,j,cnt=0;
    //      p[i]          （     -ans，     +ans）   ，       
    for(i=l; i<=r; ++i)
        if(  p[i].x>=p[mid].x-ans && p[i].x<=p[mid].x+ans  )
            arr[cnt++]=i;
    //             arr        
    sort(arr,arr+cnt,cmp_y);
    for(i=0; i<cnt; i++)
        for(j=i+1; j<cnt; j++)
        {
            if(p[arr[j]].y-p[arr[i]].y>=ans) break;
            ans=Min(ans,dis(p[arr[i]],p[arr[j]]));
        }

    return ans;
}

int main()
{
    int i,n;
    while( scanf("%d",&n)!=EOF && n)
    {
        for(i=0;i<n;++i)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        //                 
        sort(p,p+n,cmp_x);
        printf("%.2lf
",close_pair(0,n-1)/2.0);
    }
    return 0;
}