zoj 1450 Minimal Circle 最小覆盖圆

题目链接：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=450

You are to write a program to find a circle which covers a set of points and has the minimal area. There will be no more than 100 points in one problem.

题意描述：找到一个最小圆能够包含到所有的二维坐标点。

算法分析：最小覆盖圆的做法。

 //最小覆盖圆
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<cmath>
 #include<algorithm>
 #define inf 0x7fffffff
 #define exp 1e-10
 #define PI 3.141592654
 using namespace std;
 const int maxn=+;
 struct Point
 {
     double x,y;
     Point(double x=,double y=):x(x),y(y){}
 }an[maxn],d;//d:最小覆盖圆的圆心坐标
 double r;//最小覆盖圆的半径
 typedef Point Vector;
 Vector operator + (Vector A,Vector B) {return Vector(A.x+B.x , A.y+B.y); }
 Vector operator - (Vector A,Vector B) {return Vector(A.x-B.x , A.y-B.y); }
 Vector operator * (Vector A,double p) {return Vector(A.x*p , A.y*p); }
 Vector operator / (Vector A,double p) {return Vector(A.x/p , A.y/p); }
 int dcmp(double x)
 {
     if (fabs(x)<exp) return ;
     return x< ? - : ;
 }
 double cross(Vector A,Vector B)
 {
     return A.x*B.y-B.x*A.y;
 }
 double dist(Vector A,Vector B)
 {
     double x=(A.x-B.x)*(A.x-B.x);
     double y=(A.y-B.y)*(A.y-B.y);
     return sqrt(x+y);
 }

 void MiniDiscWith2Point(Point p,Point q,int n)
 {
     d=(p+q)/2.0;
     r=dist(p,q)/;
     int k;
     double c1,c2,t1,t2,t3;
     for (k= ;k<=n ;k++)
     {
         if (dist(d,an[k])<=r) continue;
         if (dcmp(cross(p-an[k],q-an[k]))!=)
         {
             c1=(p.x*p.x+p.y*p.y-q.x*q.x-q.y*q.y)/2.0;
             c2=(p.x*p.x+p.y*p.y-an[k].x*an[k].x-an[k].y*an[k].y)/2.0;
             d.x=(c1*(p.y-an[k].y)-c2*(p.y-q.y))/((p.x-q.x)*(p.y-an[k].y)-(p.x-an[k].x)*(p.y-q.y));
             d.y=(c1*(p.x-an[k].x)-c2*(p.x-q.x))/((p.y-q.y)*(p.x-an[k].x)-(p.y-an[k].y)*(p.x-q.x));
             r=dist(d,an[k]);
         }
         else
         {
             t1=dist(p,q);
             t2=dist(q,an[k]);
             t3=dist(p,an[k]);
             if (t1>=t2 && t1>=t3)
             {
                 d=(p+q)/2.0;
                 r=dist(p,q)/2.0;
             }
             else if (t2>=t1 && t2>=t3)
             {
                 d=(an[k]+q)/2.0;
                 r=dist(an[k],q)/2.0;
             }
             else
             {
                 d=(an[k]+p)/2.0;
                 r=dist(an[k],p)/2.0;
             }
         }
     }
 }
 void MiniDiscWithPoint(Point p,int n)
 {
     d=(p+an[])/2.0;
     r=dist(p,an[])/2.0;
     int j;
     for (j= ;j<=n ;j++)
     {
         if (dist(d,an[j])<=r) continue;
         else
         {
             MiniDiscWith2Point(p,an[j],j-);
         }
     }
 }

 int main()
 {
     int n;
     while (scanf("%d",&n)!=EOF && n)
     {
         for (int i= ;i<=n ;i++)
         {
             scanf("%lf%lf",&an[i].x,&an[i].y);
         }
         if (n==)
         {
             printf("%lf %lf\n",an[].x,an[].y);
             continue;
         }
         r=dist(an[],an[])/2.0;
         d=(an[]+an[])/2.0;
         for (int i= ;i<=n ;i++)
         {
             if (dist(d,an[i])<=r) continue;
             else
             MiniDiscWithPoint(an[i],i-);
         }
         printf("%.2lf %.2lf %.2lf\n",d.x,d.y,r);
     }
     return ;
 }