Quoit Design

Quoit Design

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

求一堆坐标里最短的坐标之间的距离的1/2

 

 

 

#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstdio>
using namespace std;
#define max_v 100005
int n;
struct node
{
    double x,y;
}p[max_v];
int a[max_v];
double cmpx(node a,node b)
{
    return a.x<b.x;
}
double cmpy(int a,int b)
{
    return p[a].y<p[b].y;
}
double min_f(double a,double b)
{
    return a<b?a:b;
}
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double slove(int l,int r)
{
    if(r==l+1)
        return dis(p[l],p[r]);
    if(l+2==r)
        return min_f(dis(p[l],p[r]),min_f(dis(p[l],p[l+1]),dis(p[l+1],p[r])));
    int mid=(l+r)>>1;
    double ans=min_f(slove(l,mid),slove(mid+1,r));
    int i,j,cnt=0;
    for( i=l;i<=r;i++)
    {
        if(p[i].x>=p[mid].x-ans&&p[i].x<=p[mid].x+ans)
        {
            a[cnt++]=i;
        }
    }
    sort(a,a+cnt,cmpy);
    for(i=0;i<cnt;i++)
    {
        for(j=i+1;j<cnt;j++)
        {
            if(p[a[j]].y-p[a[i]].y>=ans)
                break;
            ans=min_f(ans,dis(p[a[i]],p[a[j]]));
        }
    }
    return ans;
}
int main()
{
    int i;
    while(~scanf("%d",&n))
    {
        if(n==0)
            break;
        for(i=0;i<n;i++)
        {
            scanf("%lf %lf",&p[i].x,&p[i].y);
        }
        sort(p,p+n,cmpx);
        printf("%0.2lf\n",slove(0,n-1)/2.0);
    }
    return 0;
}