题意：求空间两线的最短距离和最短线的交点

题解：

线性代数和空间几何，主要是用叉积，点积，几何。

知道两个方向向量s1，s2，求叉积可以得出他们的公共垂直向量，然后公共垂直向量gamma和两线上的点形成的向量做内积，

在除掉gamma的长度就得到投影，即是最短距离。

然后求两个点可以用gamma和s2的叉积和l2上的一个点描述一个平面，再求平面和线的交点，

把（p2-p1）*n 和（p0-p1）*n相除算出比例乘上p2-p1得到交点和p1的差，再加上p1就求出交点了

学习点：计算几何的一些东西

 

#include
     
      #include
      
       inline double fun(double a, double b, double c, double d){    return a*d - b*c;}#define squ(x) ((x)*(x))struct Poi{    double x,y,z;    Poi(double X = 0, double Y = 0, double Z = 0){        x = X; y = Y; z = Z;    }    void input(){        scanf("%lf%lf%lf",&x,&y,&z);    }    Poi operator + (const Poi & rhs){        return Poi(x+rhs.x,y+rhs.y,z+rhs.z);    }    Poi operator - (Poi & rhs){        return Poi(x-rhs.x,y-rhs.y,z-rhs.z);    }    Poi operator ^(Poi & rhs){        return Poi(fun(y,z,rhs.y,rhs.z),-fun(x,z,rhs.x,rhs.z),fun(x,y,rhs.x,rhs.y));    }    Poi operator *(double t){        return Poi(x*t,y*t,z*t);    }    double operator *(const Poi & rhs){        return x*rhs.x+y*rhs.y+z*rhs.z;    }};typedef Poi Vector;double Dot(const Vector & a,const Poi& b) {     return a.x*b.x+a.y*b.y+a.z*b.z;}Poi LinePlaneIns(Poi &p1,Poi &p2,Poi &p0,Vector &n){    Vector v = p2 - p1;    double Ratio = (Dot(n,p0-p1))/(Dot(n,v));//保证相交    return p1+v*Ratio;}double Length(const Vector &x){    return sqrt(squ(x.x)+squ(x.y)+squ(x.z));}int main(){   // freopen("in.txt","r",stdin);    int T;    scanf("%d",&T);    while(T--){        Poi p1,p2,p3,p4;        p1.input();        p2.input();        p3.input();        p4.input();        Vector alpha = p2 - p1;        Vector beta = p4 - p3;        Vector gamma = alpha^beta;        double distance = fabs(Dot(gamma,(p1-p3))/Length(gamma));        Vector n1 = alpha^gamma;        Vector n2 = beta^gamma;        Poi ins1 = LinePlaneIns(p1,p2,p3,n2);        Poi ins2 = LinePlaneIns(p3,p4,p1,n1);        printf("%.6lf\n%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n",distance,ins1.x,ins1.y,ins1.z,ins2.x,ins2.y,ins2.z);    }    return 0;}