也就是已知空间不共圆的四点坐标,怎么求这个四点所组成的外接球的半径???
也就是已知空间不共圆的四点坐标,怎么求这个四点所组成的外接球的半径???
 2006-03-07 17:34
	    2006-03-07 17:34
  太深奥了
 2006-03-09 22:18
	    2006-03-09 22:18
  这个问题,我解答出来了,但是没有很多的测试数据,所以不知道对不对……
//main.cpp
#include "classes.h"
void main()
{
  Point a;
  Point b;
  Point c;
  Point d;
  Equation m (a,b);
  Equation n (a,c);
  Equation l (a,d);
  answer (m,l,n);
}
//classes.cpp
#include "classes.h"
#include <iostream.h>
Point::Point ()
{
    cout <<"Please input a point:";     //输入一个点的坐标
 cin  >>x>>y>>z;
}
float Point::getX ()
{
 return x;
}
float Point::getY ()
{
 return y;
}
float Point::getZ ()
{
 return z;
}
Equation::Equation (Point a,Point b)
{
 float i,j,k;
    
 i=(a.getX()+b.getX())/2;    //中点
 j=(a.getY()+b.getY())/2;
 k=(a.getZ()+b.getZ())/2;
 
 x=(a.getX()-b.getX());      //向量
 y=(a.getY()-b.getY());
 z=(a.getZ()-b.getZ());
 
 t=i*x+j*y+k*z;              //方程 各未知数系数即x,y,z   t为方程右边常数
}
void answer (Equation a,Equation b,Equation c) 
{
 int i,j;
 float x,y,z,D=0,E=0;
 float p[3][4]={{a.x,a.y,a.z,a.t},{b.x,b.y,b.z,b.t},{c.x,c.y,c.z,c.t}};
 
 for (i=0;i<=2;i++)
  D+=p[0][i]*p[1][(i+1)%3]*p[2][(i+2)%3];
 for (i=0;i<=2;i++)
  D-=p[2][i]*p[1][(i+1)%3]*p[0][(i+2)%3];
 
 for (i=0;i<=2;i++)              //x
 {
  j=p[i][0];
  p[i][0]=p[i][3];
  p[i][3]=j;
 }
 for (i=0;i<=2;i++)
  E+=p[0][i]*p[1][(i+1)%3]*p[2][(i+2)%3];
 for (i=0;i<=2;i++)
  E-=p[2][i]*p[1][(i+1)%3]*p[0][(i+2)%3];
    x=E/D;
    for (i=0;i<=2;i++)
 {
  j=p[i][0];
  p[i][0]=p[i][3];
  p[i][3]=j;
 }
    
 E=0;                            //y
 for (i=0;i<=2;i++)
 {
  j=p[i][1];
  p[i][1]=p[i][3];
  p[i][3]=j;
 }
 for (i=0;i<=2;i++)
  E+=p[0][i]*p[1][(i+1)%3]*p[2][(i+2)%3];
 for (i=0;i<=2;i++)
  E-=p[2][i]*p[1][(i+1)%3]*p[0][(i+2)%3];
    y=E/D;
    for (i=0;i<=2;i++)
 {
  j=p[i][1];
  p[i][1]=p[i][3];
  p[i][3]=j;
 }
    E=0;                            //z
 for (i=0;i<=2;i++)
 {
  j=p[i][2];
  p[i][2]=p[i][3];
  p[i][3]=j;
 }
 for (i=0;i<=2;i++)
  E+=p[0][i]*p[1][(i+1)%3]*p[2][(i+2)%3];
 for (i=0;i<=2;i++)
  E-=p[2][i]*p[1][(i+1)%3]*p[0][(i+2)%3];
    z=E/D;
    for (i=0;i<=2;i++)
 {
  j=p[i][2];
  p[i][2]=p[i][3];
  p[i][3]=j;
 }
    cout <<"The point is "<<x<<","<<y<<","<<z<<"\n";
}
//classes.h
class Point
{
public:
    Point ();
 float getX();
    float getY();
    float getZ();
private:
    float x;
 float y;
 float z;
};
class Equation
{
public:
    Equation (Point a,Point b);
    friend void answer (Equation a,Equation b,Equation c);
private:
 float x;
 float y;
 float z;
 float t;
};
还请高手指正。
 2006-03-15 18:50
	    2006-03-15 18:50
  谢谢,我用简单数据进行测试,到时可以算出答案!
[此贴子已经被作者于2006-3-21 17:42:20编辑过]

 2006-03-21 16:06
	    2006-03-21 16:06
   2006-04-15 15:52
	    2006-04-15 15:52