HDU2215 ——Maple trees

Problem Description

There are a lot of trees in HDU. Kiki want to surround all the trees with the minimal required length of the rope . As follow,

To make this problem more simple, consider all the trees are circles in a plate. The diameter of all the trees are the same (the diameter of a tree is 1 unit). Kiki can calculate the minimal length of the rope , because it's so easy for this smart girl.
But we don't have a rope to surround the trees. Instead, we only have some circle rings of different radius. Now I want to know the minimal required radius of the circle ring. And I don't want to ask her this problem, because she is busy preparing for the examination.
As a smart ACMer, can you help me ?

 

Input

The input contains one or more data sets. At first line of each input data set is number of trees in this data set n （1 <= n <= 100）, it is followed by n coordinates of the trees. Each coordinate is a pair of integers, and each integer is in [-1000, 1000], it means the position of a tree’s center. Each pair is separated by blank.
Zero at line for number of trees terminates the input for your program.

 

Output

Minimal required radius of the circle ring I have to choose. The precision should be 10^-2.

 

Sample Input

     
     
     
     

      
      
      
      2
1 0
-1 0
0
     
     
     
     

 

Sample Output

     
     
     
     

      
      
      
      1.50
     
     
     
     

 

Author

zjt

 

计算几何题，最小圆覆盖，这里我用的是随机增量法，期望为O(n)。

    先以第一个点为圆心，找到第一个不在圆内或圆上的点，以它为圆心，去找第二个不在圆内或圆上的点，以这两个点的中点为圆心，距离为直径，去找第三个不在圆上或者圆内的点，以这三个点形成的三角形的外心为圆心，遍历其他点，最后得到的就是覆盖所有点的最小的圆。

//最小圆覆盖——随机增量法，期望为O(n)

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<map>
#include<vector>
#include<math.h>

using namespace std;

const double eps = 1e-8;
struct point
{
    double  x ,y;
}p[105];

double dist(point a, point b)
{
    return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));//计算距离
}

//返回外心
point crosspoint(point p0, point p1, point p2)
{
    point ret;
    double a1 = p1.x - p0.x, b1 = p1.y - p0.y, c1 = (a1 * a1 + b1 * b1) / 2;
    double a2 = p2.x - p0.x, b2 = p2.y - p0.y, c2 = (a2 * a2 + b2 * b2) / 2;
    double d = a1 * b2 - a2* b1;
    ret.x = p0.x + (c1 * b2 - c2 * b1) / d;
    ret.y = p0.y + (a1 * c2 - a2 * c1) / d;
    return ret;
}

double min_cover_circle(int n)
{
    random_shuffle(p, p + n) ;//打乱顺序
    point c = p[0];
    double r = 0;
    for (int i = 1; i < n; ++i)
    {
        if (dist(p[i], c) > r) //第一个点
        {
            c = p[i];
            r = 0;
            for (int j = 0; j < i; ++j)
            {
                if (dist(p[j], c) > r)
                {
                    c.x = (p[i].x + p[j].x) / 2;
                    c.y = (p[i].y + p[j].y) / 2;
                    r=dist(p[j], c);
                    for (int k = 0; k < j; ++k)
                    {
                        if(dist(p[k], c) > r)
                        {
                            c = crosspoint(p[i], p[j], p[k]);
                            r = dist(c, p[k]);
                        }
                    }
                }
            }
        }
    }
    return r;
}

int main()
{
    int n;
    while(~scanf("%d",&n) && n)
    {
        for (int i = 0; i < n; ++i)
        {
            scanf("%lf%lf",&p[i].x ,&p[i].y);
        }
        if(n == 1)
            printf("0.50\n");//这题树本身也存在半径，而计算时用的是中心，所以输出时还要加上树的半径
        else
        {
            double r = min_cover_circle(n);
            printf("%.2lf\n", r + 0.50);
        }
    }
    return 0;
}