HDU 1007.Quoit Design【投机取巧（枚举）】【9月9】

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
   
   
   
   

 

求最近的两点之间距离的一半。应该用分治法求的，但是这个道题数据有点弱，安x坐标排序，只需枚举附近3个点之间的距离就可以AC，但是时间很长。至于分治法，还没懂。下次补充。代码如下：

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
struct ss{
    double x,y;
};
bool cmp(ss a,ss b){
    if(a.x<b.x) return true;
    else return false;
}
int main()
{
    int T;
    while(scanf("%d",&T)==1&&T){
        ss f[100010];
        for(int i=0;i<T;i++)
            scanf("%lf%lf",&f[i].x,&f[i].y);
        sort(f,f+T,cmp);
        double len=999999999;
        for(int i=0;i<T;i++)
        for(int j=i+1;j<i+4&&j<T;j++){
            double lenx=sqrt((f[i].x-f[j].x)*(f[i].x-f[j].x)+(f[i].y-f[j].y)*(f[i].y-f[j].y));
            if(lenx<len) len=lenx;
        }
        printf("%.2f\n",len/2);
    }
    return 0;
}