hdu1007Quoit 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

20 01 121 11 13-1.5 00 00 1.50

 

Sample Output

0.710.000.75

/*题意：给出一堆平面上的点，求最近点对。做法：用结构体存每个点的x、y的值。      ①将每个点 x 值由小到大进行排序，若 x 相同，则按照 y 由小到大进行排序。      ②将每个点 y 值由小到大进行排序，若 y 相同，则按照 x 由小到大进行排序。      分别用一层循环不停的更新最小距离*/#include <cstdio>#include <cstring>#include <cmath>#include <iostream>#include <algorithm>#include <string>#include <cstdlib>using namespace std;const int maxn = 100000+10;struct node{double x,y;}p[maxn];bool cmp1(const node a,const node b){    if(a.x==b.x) return a.y<b.y;    return a.x<b.x;}bool cmp2(const node a,const node b){    if(a.y==b.y) return a.x<b.x;    return a.y<b.y;}double work(const node a,const node b){    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));}int main(void){    int n;    while(scanf("%d",&n)!=EOF,n)    {        double mi;        for(int i=0;i<n;i++)            scanf("%lf%lf",&p[i].x,&p[i].y);        sort(p,p+n,cmp1);        mi = 100000000.0;        for(int i=0;i<n-1;i++)            mi=min(work(p[i],p[i+1]),mi);        sort(p,p+n,cmp2);        for(int i=0;i<n-1;i++)            mi=min(work(p[i],p[i+1]),mi);        printf("%.2f\n",mi/2);    }    return 0;}