tetrahedron (公式)

Posted 2021-01-10 wethura

我是直接搬运了某大佬的代码，毕竟我不清楚如何计算这个东西。

其中四点共面的求法就是体积为0，然后圆心和半径就公式了。

#include<cstdio>
#include<iostream>
#include<cmath>
using namespace std;
#define eps 1e-8
struct node
{
    double x,y,z;
}a,b,c,d,ans;

///面积公式
double get(node a,node b,node c)
{
    double x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
    double x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
    double d1=y1*z2-y2*z1;
    double d2=x1*z2-x2*z1;
    double d3=x1*y2-x2*y1;
    return sqrt(d1*d1+d2*d2+d3*d3)*0.5;
}

///体积
double Get(node a,node b,node c,node d)
{
    double x1=b.x-a.x,y1=b.y-a.y,z1=b.z-a.z;
    double x2=c.x-a.x,y2=c.y-a.y,z2=c.z-a.z;
    double x3=d.x-a.x,y3=d.y-a.y,z3=d.z-a.z;
    double ans=x1*(y2*z3-y3*z2)-y1*(x2*z3-x3*z2)+z1*(x2*y3-x3*y2);
    return abs(ans)/6.0;
}
int main()
{
    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,s2,s3,s4,V;
        V=Get(a,b,c,d);
        s1=get(b,c,d);
        s2=get(a,c,d);
        s3=get(a,b,d);
        s4=get(a,b,c);
        double t=s1+s2+s3+s4;
        if(V<eps)
        {
            printf("O O O O
");
            continue;
        }
        ans.x=(s1*a.x+s2*b.x+s3*c.x+s4*d.x)/t;
        ans.y=(s1*a.y+s2*b.y+s3*c.y+s4*d.y)/t;
        ans.z=(s1*a.z+s2*b.z+s3*c.z+s4*d.z)/t;
        double r=V*3.0/t;
        printf("%.4lf %.4lf %.4lf %.4lf
",ans.x,ans.y,ans.z,r);
    }
}