Journey to Jupiter（三维计算几何）

原题： https://cn.vjudge.net/problem/Gym-101991J

题意：

有一个正三边形的平面（中心为(0,0,0)），现在告诉你一个顶点A的坐标和平面法向量alpha，边长，问三个点离H(0,0,-h)的距离

解析：

用O - alpha和O - A叉积，得到一个平行与BC的向量，那么很容易得到OB和OC，BH=OH-OB，CH=OH-OC，就解出来了

#include<bits/stdc++.h>
using namespace std;
const double eps=1e-8;
const double pi=acos(-1);
int dcmp(double x){
    if(abs(x)<eps) return 0;
    if(x>0) return 1;
    return -1;
}
struct Point{
    double x,y,z;
    Point(){}
    Point(double x,double y,double z):x(x),y(y),z(z){}
};
typedef Point Vector;
bool operator == (const Point &a, const Point &b){
  return dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0 &&dcmp(a.z - b.z) == 0;
}
Vector operator + (Vector a, Vector b){
    return Vector(a.x + b.x, a.y + b.y,a.z + b.z);
}
Vector operator - (Vector a, Vector b){
    return Vector(a.x - b.x, a.y - b.y,a.z - b.z);
}
Vector operator * (Vector a, double p){
    return Vector(a.x*p, a.y*p,a.z*p);
}
Vector operator / (Vector a, double p){
    return Vector(a.x / p, a.y / p,a.z/p);
}
Vector Across(Vector a,Vector b){
    return Vector(a.y*b.z-b.y*a.z,b.x*a.z-a.x*b.z,a.x*b.y-b.x*a.y);
}

double Dot(Vector a, Vector b){//内积
    return a.x*b.x + a.y*b.y+a.z*b.z;
}
double Length(Vector a){
    return sqrt(Dot(a,a));
}

double Distance(Point a, Point b){//两点间距离
    return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y)+(a.z-b.z)*(a.z-b.z));
}



int main(){
    freopen("jupiter.in","r",stdin);
    int t;scanf("%d",&t);
    while(t--){

        double xx,yy,zz;
        scanf("%lf%lf%lf",&xx,&yy,&zz);

        Vector alpha(xx,yy,zz);

        scanf("%lf%lf%lf",&xx,&yy,&zz);
        Vector A(xx,yy,zz),O(0,0,0);
        double Len,H;
        scanf("%lf%lf",&Len,&H);

        Vector OA = A-O , opOA = O-A;
        opOA=opOA/Length(opOA);
        opOA=opOA*Len*sqrt(3)/6.0;

        Vector OB =  Across(alpha,A-O);
        OB=OB/Length(OB);
        OB=OB*Len/2;

        Vector OC = OB * -1.0;

        OB=OB+opOA;
        OC=OC+opOA;

        Vector OH {0,0,-H};

        double oa=Length(OH-OA),ob=Length(OH-OB),oc=Length(OH-OC);
        printf("%.6f %.6f %.6f\n",oa,ob,oc);
    }
}