Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 33577    Accepted Submission(s): 8800

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

存点的下标比存点快得多、- -

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
using namespace std;
#define INF 1e20
#define N 100010

struct Point
{
    double x,y;
};

int n;
Point p[N];
int tmp[N];

double dis(int a,int b)
{
    return sqrt((p[a].x-p[b].x)*(p[a].x-p[b].x)+(p[a].y-p[b].y)*(p[a].y-p[b].y));
}
bool cmp(Point a,Point b)
{
    if(a.x!=b.x)return a.x<b.x;
    return a.y<b.y;
}
bool cmpy(int a,int b)
{
    return p[a].y<p[b].y;
}

double solve(int l,int r)
{
    if(l==r) return INF;               //一个点的情况
    if(l+1==r) return dis(l,r);        //两个点的情况
    int m=(l+r)>>1;
    double d1=solve(l,m);
    double d2=solve(m+1,r);
    double d=min(d1,d2);
    int k=0;
    for(int i=l;i<=r;i++)              //分离出宽度为d的区间
    {
        if(fabs(p[m].x-p[i].x)<=d)
            tmp[++k]=i;
    }
    sort(tmp+1,tmp+k+1,cmpy);
    for(int i=1;i<=k;i++)              //线性扫描 ,判断是否有小于d的距离，由于是按y排序的，所以省去了不少步骤
    {
        for(int j=i+1;j<=k && p[tmp[j]].y-p[tmp[i]].y<d;j++)
        {
            d=min(d,dis(tmp[i],tmp[j]));
        }
    }
    return d;
}
int main()
{
    while(scanf("%d",&n) && n)
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%lf%lf",&p[i].x,&p[i].y);
        }
        sort(p+1,p+n+1,cmp);
        printf("%.2f\n",solve(1,n)/2);
    }
    return 0;
}