hdu 1007 Quoit Design (最近点对问题)

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25335    Accepted Submission(s): 6716

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

 

 

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//1468MS    2288K    1378 B    G++
/*

    题意；
        给出n个点，求最近的两个点的距离的一般
        
    最近点对：
        要用到分治的思想做，是时间复杂度为O(nlgn)，直接求会超时
        最近点对问题在很多算法书和资料上都有讲到，主要思想是分治。 
    
    注意一点： 用C的 qsort排序一直TLE，表示有点无力.. 

*/
#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<math.h>
#define N 100005
using namespace std;
struct node{
    double x,y;       
}p[N];
int a[N];
int n;
double cmpx(node a,node b)
{
    return a.x<b.x;
}
double cmpy(int a,int b)
{
    return p[a].y<p[b].y;
}
inline double Max(double a,double b)
{
    return a>b?a:b;
}
inline double Min(double a,double b)
{
    return a<b?a:b;
}
inline 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 find(int l,int r)
{
    if(l==r) return 0x7ffffff; 
    if(r==l+1) return Dis(p[l],p[r]);
    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 d=Min(find(l,mid),find(mid+1,r));
    int cnt=0;
    for(int i=l;i<=r;i++){
        if(p[i].x>=p[mid].x-d && p[i].x<=p[mid].x+d)
            a[cnt++]=i;
    }
    sort(a,a+cnt,cmpy);
    for(int i=0;i<cnt;i++){
        for(int j=i+1;j<cnt;j++){
            if(p[a[j]].y-p[a[i]].y>=d) break;
            d=Min(d,Dis(p[a[i]],p[a[j]]));
        }
    }
    return d;
}
int main(void)
{
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0) break;
        for(int i=0;i<n;i++)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        sort(p,p+n,cmpx);
        printf("%.2lf\n",find(0,n-1)/2);
    }
    return 0;
}