最近对问题，分治法求解

最近对问题，即从一堆点中，找出离得最近的两个点之间的距离

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
   
   
   
   

<span style="font-family:Courier New;">#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn=100000+5;
const int INF=1<<31-1;
struct Point
{
    double x,y;
};

Point p[maxn];
Point pt[maxn];

bool cmp(Point &p1,Point &p2)
{
    return p1.x<p2.x;
}
bool cmp2(Point &p1,Point &p2)
{
    return p1.y<p2.y;
}
double getDis(Point &p1,Point &p2)
{
    return sqrt( pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2) );
}
double getClosest(int left,int right)
{
    if(left==right) return INF;
    if(right-left==1) return getDis(p[left],p[right]);
    int mid=(left+right)/2;
    double d1=getClosest(left,mid);
    double d2=getClosest(mid+1,right);
    double d=(d1<d2?d1:d2);
    int index=0;
    for(int i=mid;i>=left;i--)
    {
        if(p[mid].x-p[i].x>d) break;
        else pt[index++]=p[i];
    }
    for(int i=mid+1;i<=right;i++)
    {
        if(p[i].x-p[mid].x>d) break;
        else pt[index++]=p[i];
    }
    sort(pt,pt+index,cmp2); //这里一定要排序，排序后，按照鸽笼原理，dis最多更新8次，可减少计算
    for(int i=0;i<index-1;i++)
    {
        for(int j=i+1;j<index;j++)
        {
            if(pt[j].y-pt[i].y>d) break;
            else
            {
                double dis=getDis(pt[i],pt[j]);
                if(dis<d) d=dis;
            }
        }
    }
    return d;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF && n)
    {
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        sort(p,p+n,cmp);
        double ans=getClosest(0,n-1);
        printf("%.2lf\n",ans/2);
    }
    return 0;
}</span>