增量法最小圆覆盖....

Maple trees

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1646 Accepted Submission(s): 510

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 rZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vcGUgLCBiZWNhdXNlIGl0"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

Sample Output

1.50

Author zjt

/* ***********************************************
Author        :CKboss
Created Time  :2014年12月29日 星期一 17时19分19秒
File Name     :HDOJ2215.cpp
************************************************ */

#include 
  
   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
             using namespace std; const int maxn = 111; const double eps=1e-8; int dcmp(double x) { if(fabs(x)
             
              eps?1:-1; } struct Point { double x,y; Point(){} Point(double _x,double _y) { x=_x; y=_y;} }pt[maxn]; struct Circle { Point c; double r; Circle(){} Circle(Point _c,double _r) { c=_c; r=_r; } }; Point operator+(Point A,Point B) { return Point(A.x+B.x,A.y+B.y); } Point operator-(Point A,Point B) { return Point(A.x-B.x,A.y-B.y); } Point operator*(Point A,double p) { return Point(A.x*p,A.y*p); } Point operator/(Point A,double p) { return Point(A.x/p,A.y/p); } double Dot(Point A,Point B) { return A.x*B.x+A.y*B.y; } double Length(Point A) { return sqrt(Dot(A,A)); } double Cross(Point A,Point B) { return A.x*B.y-A.y*B.x; } Circle CircumscribedCircle(Point p1,Point p2,Point p3) { double Bx=p2.x-p1.x,By=p2.y-p1.y; double Cx=p3.x-p1.x,Cy=p3.y-p1.y; double D=2*(Bx*Cy-By*Cx); double cx=(Cy*(Bx*Bx+By*By)-By*(Cx*Cx+Cy*Cy))/D+p1.x; double cy=(Bx*(Cx*Cx+Cy*Cy)-Cx*(Bx*Bx+By*By))/D+p1.y; Point p=Point(cx,cy); return Circle(p,Length(p1-p)); } void min_cover_circle(Point p[],int n,Circle& c) { c.c=p[0]; c.r=0; for(int i=1;i
              
               0) { c.c=p[i]; c.r=0; for(int j=0;j
               
                0) { c.c=Point((p[i].x+p[j].x)/2.,(p[i].y+p[j].y)/2.); c.r=Length(p[j]-p[i])/2.; for(int k=0;k
                
                 0) { c=CircumscribedCircle(p[i],p[j],p[k]); } } } } } } } int n; int main() { //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(scanf("%d",&n)!=EOF&&n) { for(int i=0;i

HDOJ 2215 Maple trees 最小圆覆盖

Maple trees