hdu 5733 tetrahedron 求四面体内切球球心坐标及半径大小

tetrahedron

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 720    Accepted Submission(s): 291

Problem Description

Given four points ABCD, if ABCD is a tetrahedron, calculate the inscribed sphere of ABCD.

 

Input

Multiple test cases (test cases ≤100).

Each test cases contains a line of 12 integers [−1e6,1e6] indicate the coordinates of four vertices of ABCD.

Input ends by EOF.

 

Output

Print the coordinate of the center of the sphere and the radius, rounded to 4 decimal places.

If there is no such sphere, output "O O O O".

 

Sample Input

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

 

Sample Output

0.4226 0.4226 0.4226 0.4226O O O O

 

Author

HIT

 

Source

2016 Multi-University Training Contest 1

 

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    题意：给出四个点，要你求四面体内切球球心坐标及半径大小。无解输出"O O O O"。

     解：首先如果四点共面，必然无解。(我的判断方法是：ABC三点组成的平面法向量n和向量AD如果点乘为0，那么四点共面)

         否则，必然有解(四面体必有内切球，可以用分角面证明)

内切球球心(x,y,z)的求法是

  令  S=   S△ABC+S△ACD+S△ABD+S△BCD;

   x=    (S△ABC*D.x+S△ACD*B.x+S△ABD*C.x+S△BCD*A.x)/S;

   y=    (S△ABC*D.y+S△ACD*B.y+S△ABD*C.y+S△BCD*A.y)/S;

   z=    (S△ABC*D.z+S△ACD*B.z+S△ABD*C.z+S△BCD*A.z)/S;

半径大小R可以根据 方程   1.0/4*S△ABC*H= 1.0/4* S*R   (依据四面体体积相等) 解得

细节上，求三角形面积用了海伦公式，求ABC面上的高H用了H= |      向量AB*法向量n/| 法向量n |     |

   

#include<cstdio>#include<string>#include<cstring>#include<iostream>#include<cmath>#include<algorithm>#include<vector>using namespace std;#define all(x) (x).begin(), (x).end()#define for0(a, n) for (int (a) = 0; (a) < (n); (a)++)#define for1(a, n) for (int (a) = 1; (a) <= (n); (a)++)#define sqr(x)  ((x)*(x))const double eps=1e-7;typedef long long ll;typedef pair<int, int> pii;const int INF =0x3f3f3f3f;struct Point{   double x,y,z;   Point(){}   Point(double x,double y,double z):x(x),y(y),z(z){}}A,B,C,D,O;typedef Point Vector;double R;Vector operator-(const Point &A,const Point &B){    return Vector(A.x-B.x,A.y-B.y,A.z-B.z);}Vector operator-(const Point &A){    return Vector(-A.x,-A.y,-A.z);}double Dot(Vector A,Vector B){    return (A.x*B.x+A.y*B.y+A.z*B.z);}Vector Cross(Vector a,Vector b){    return Vector( (a.y*b.z-a.z*b.y),-(a.x*b.z-b.x*a.z),(a.x*b.y-a.y*b.x) );}double Length(Vector a){    return sqrt( Dot(a,a));}double Helen(Point A,Point B,Point C)//海伦公式{     Vector AB=B-A;double c= Length(AB);     Vector AC=C-A;double b= Length(AC);     Vector BC=C-B;double a= Length(BC);     double p=(a+b+c)/2;     return sqrt( p*(p-a)*(p-b)*(p-c) );}bool solve(Point A,Point B,Point C,Point D){   Vector AB=B-A;   Vector BC=C-B;   Vector n= Cross(AB,BC);          //法向量   double ret=Dot(D-A,n);   if ( fabs(ret)<eps ) return false;//四点共面   double S=Helen(A,B,C  );   O.x+=   S*D.x;   O.y+=   S*D.y;   O.z+=   S*D.z;   return true;}void getRandXYZ(){    Vector AB=B-A;    Vector BC=C-B;    Vector n= Cross(AB,BC);    Vector AD=D-A;    if(Dot(AD,n)<0) n=-n;    double h=  Dot(AD,n)/Length(n);    double V4= h*Helen(A,B,C);    double s=0;    s+=Helen(A,B,C);    s+=Helen(A,B,D);    s+=Helen(A,C,D);    s+=Helen(B,C,D);    R=V4/s;//半径求得    O.x/=s;//球心坐标求得    O.y/=s;    O.z/=s;}int main(){    while(~scanf("%lf%lf%lf",&A.x,&A.y,&A.z))    {        scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf",&B.x,&B.y,&B.z,&C.x,&C.y,&C.z,&D.x,&D.y,&D.z );        memset(&O,0,sizeof O);        if(!solve(A,B,C,D) )        {            puts("O O O O");            continue;        }        solve(A,B,D,C);        solve(A,C,D,B);        solve(B,C,D,A);        getRandXYZ();        printf("%.4f %.4f %.4f %.4f\n",O.x,O.y,O.z,R);     }   return 0;}