Quoit Design

Quoit Design

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46340    Accepted Submission(s): 12084

这题可以分治：

首先按照x方向排序，然后分别对 两边求最短距离。

然后按照y排序，统计距离分割点-d 和+d之间的点，算了，忘了。

但是可以直接排序，然后枚举相邻三个点之间到距离。这三个点的组合总计三个距离，可以证明最短到距离一定存在于某个

三元组里。

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

 

Recommend

JGShining

 

#include<iostream>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
const int maxx=100005;
struct Node{
    double x;
    double y;
}node[maxx];
bool cmp(Node a,Node b){
    if(a.x!=b.x) return a.x<b.x;
    return a.y<b.y;
}
double dis(Node a,Node b){
    double dx,dy;
    dx=a.x-b.x;
    dy=a.y-b.y;
    double ans=dx*dx+dy*dy;
    return sqrt(ans);
}
double min(double a,double b,double c){

    double tmp=0.0;
    tmp=(a<=b)?a:b;
    tmp= (tmp<=c)?tmp:c;
    //cout<<a<<"  "<<b<<" "<<c<<"  "<<tmp<<endl;
    return tmp;
}
int main(){
        
    int n;
    while(scanf("%d",&n)&&n!=0){

        for(int i=0;i<n;i++){
            scanf("%lf%lf",&node[i].x,&node[i].y);
        }
        sort(node,node+n,cmp);
        double ans=1000000.0;
        if(n==1){
            printf("0.00
");
            continue;
        }else if(n==2){
            printf("%.2lf
",dis(node[0],node[1])/2);
            continue;
        }
        double tmp;
        for(int i=1;i<n-1;i++){
            tmp=min(dis(node[i-1],node[i]),dis(node[i],node[i+1]),dis(node[i-1],node[i+1]));
            ans=min(ans,tmp);
        }
        ans=ans/2;
        printf("%.2lf
",ans);
    }
    return 0;
}