hdu 5733

http://acm.split.hdu.edu.cn/showproblem.php?pid=5733

题意：给你4个定点，这个4个定点构成一个四边形，求这个四边形的内切圆的圆心和半径

没有的话输出O O O O

思路：混合积可以求四边形的体积

　　r = v*3/(s1+s2+s3+s4)

　　s1,s2,s3,s4分别是四个面的面积

然后圆心

　　x = a.x*s1+b.x*s2+c.x*s3+d.x*s4;
　　y = a.y*s1+b.y*s2+c.y*s3+d.y*s4;
　　z = a.z*s1+b.z*s2+c.z*s3+d.z*s4;

s1 是a点不构成的面积

s2 是b点不构成的面积

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <iostream>
using namespace std;

struct point3
{
    double x,y,z;
};

double lenth(point3 u){
    return sqrt(u.x*u.x+u.y*u.y+u.z*u.z);
}
point3 xmult(point3 u,point3 v)
{
    point3 ret;
    ret.x=u.y*v.z-v.y*u.z;
    ret.y=u.z*v.x-u.x*v.z;
    ret.z=u.x*v.y-u.y*v.x;
    return ret;
}

double dmult(point3 u,point3 v)
{
    return u.x*v.x+u.y*v.y+u.z*v.z;
}


point3 operator - ( point3 A , point3 B )
{
    point3 ret;
    ret.x = A.x-B.x;
    ret.y = A.y-B.y;
    ret.z = A.z-B.z;
    return ret;
}
double area( point3 A , point3 B , point3 C )
{
    return lenth(xmult(B-A,C-A))/2 ;
}

double volume( point3 A , point3 B , point3 C , point3 D )
{
   // printf("%lf
",xmult(B-A,C-A));
    //printf("%lf
",dmult(D-A,xmult(B-A,C-A)));
    return fabs(dmult(D-A,xmult(B-A,C-A)))/6 ;
}



int main()
{
    55     point3 a,b,c,d;
    while(~scanf("%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf",&a.x,&a.y,&a.z,&b.x,&b.y,&b.z,&c.x,&c.y,&c.z,&d.x,&d.y,&d.z))
    {
        double s1 = area(b,c,d);
        double s2 = area(a,c,d);
        double s3 = area(a,b,d);
        double s4 = area(a,b,c);
        double v = volume(a,b,c,d);
        if(v<1e-8)
        {
            printf("O O O O
");
            continue;
        }
        double s = s1+s2+s3+s4;
        double r = v*3/s;
        double x = a.x*s1+b.x*s2+c.x*s3+d.x*s4;
        double y = a.y*s1+b.y*s2+c.y*s3+d.y*s4;
        double z = a.z*s1+b.z*s2+c.z*s3+d.z*s4;
        printf("%.4lf %.4lf %.4lf %.4lf
",x/s,y/s,z/s,r);
    }
    return 0;
}