tetrahedron

题意：

求解一个四面体的内切球。

解法：

首先假设内切球球心为$(x0,x1,x2)$，可以用$r = \frac{3V}{S_1+S_2+S_3+S_4}$得出半径，

这样对于四个平面列出三个方程，解得

$x_n = \sum_{i=0}^3{Ai_{x_n} \cdot S_i } / (S_1 + S_2 + S_3 + S_4)$

这样，即可得出内切球。

时间复杂度$O(1)$。

 #include <iostream>

 #include <cstdio>

 #include <cstring>

 #include <cmath>

 #define LD double

 #define sqr(x) ((x)*(x))

 #define eps 1e-13

 using namespace std;

 struct node

 {

     LD x,y,z;

     void scan()

     {

         scanf("%lf%lf%lf",&x,&y,&z);

     }

     void print()

     {

         printf("%.4lf %.4lf %.4lf\n",x,y,z);

     }

     LD length()

     {

         return sqrt(sqr(x)+sqr(y)+sqr(z));

     }

     node operator+(const node &tmp)

     {

         return (node){x+tmp.x,y+tmp.y,z+tmp.z};

     }

     node operator-(const node &tmp)

     {

         return (node){x-tmp.x,y-tmp.y,z-tmp.z};

     }

     node operator/(LD tmp)

     {

         return (node){x/tmp,y/tmp,z/tmp};

     }

     node operator*(LD tmp)

     {

         return (node){x*tmp,y*tmp,z*tmp};

     }

 };

 node cross(node a,node b)

 {

     node ans;

     ans.x = a.y*b.z - b.y*a.z;

     ans.y = b.x*a.z - a.x*b.z;

     ans.z = a.x*b.y - b.x*a.y;

     return ans;

 }

 LD dist(node a,node b)

 {

     return (b-a).length();

 }

 LD dot(node a,node b)

 {

     return a.x*b.x + a.y*b.y + a.z*b.z;

 }

 LD get_angle(node a,node b)

 {

     LD tmp = dot(a,b)/a.length()/b.length();

     return acos(tmp);

 }

 node get_node(node A,node B,node C)

 {

     LD Lth = (B-A).length() + (C-A).length() + (C-B).length();

     cout << sqr(Lth-) << endl;

     LD r = fabs(cross(B-A,C-A).length()) / Lth;

     cout << r*r << endl;

     node v1 = C-A;

     node v2 = B-A;

     node v = (v1+v2)/(v1+v2).length();

     LD d = (C-A).length()/;

     LD L = sqrt(sqr(d)+sqr(r));

     v = v*L;

     return A+v;

 }

 int main()

 {

     node A,B,C,D;

     while(~scanf("%lf%lf%lf",&A.x,&A.y,&A.z))

     {

         B.scan();

         C.scan();

         D.scan();

         if(fabs(dot(cross(B-A,C-A),D-A)) < eps)

         {

             puts("O O O O");

             continue;

         }

         LD S1 = fabs(cross(B-D,C-D).length())/;

         LD S2 = fabs(cross(D-A,C-A).length())/;

         LD S3 = fabs(cross(B-A,D-A).length())/;

         LD S4 = fabs(cross(B-A,C-A).length())/;

         LD Ve = fabs(dot(cross(B-A,C-A),D-A))/;

         LD R = *Ve / ((S1+S2+S3+S4));

         node ans;

         ans.x = (S1*A.x + S2*B.x + S3*C.x + S4*D.x)/(S1+S2+S3+S4);

         ans.y = (S1*A.y + S2*B.y + S3*C.y + S4*D.y)/(S1+S2+S3+S4);

         ans.z = (S1*A.z + S2*B.z + S3*C.z + S4*D.z)/(S1+S2+S3+S4);

         printf("%.4lf %.4lf %.4lf %.4lf\n",ans.x,ans.y,ans.z,R);

     }

     return ;

 }

 /*

 0 0 0 2 0 0 0 0 2 0 2 0

 0 0 0 2 0 0 3 0 0 4 0 0

 */

巴特西

tetrahedron

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