hdu 4741 Save Labman No.004异面直线间的距离既构成最小距离的两个端点

Save Labman No.004

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1473    Accepted Submission(s): 484

Problem Description

Due to the preeminent research conducted by Dr. Kyouma, human beings have a breakthrough in the understanding of time and universe. According to the research, the universe in common sense is not the only one. Multi World Line is running simultaneously. In simplicity, let us use a straight line in three-dimensional coordinate system to indicate a single World Line.

During the research in World Line Alpha, the assistant of Dr. Kyouma, also the Labman No.004, Christina dies. Dr. Kyouma wants to save his assistant. Thus, he has to build a Time Tunnel to jump from World Line Alpha to World Line Beta in which Christina can be saved. More specifically, a Time Tunnel is a line connecting World Line Alpha and World Line Beta. In order to minimizing the risks, Dr. Kyouma wants you, Labman No.003 to build a Time Tunnel with shortest length.

 

Input

The first line contains an integer T, indicating the number of test cases.

Each case contains only one line with 12 float numbers (x1, y1, z1), (x2, y2, z2), (x3, y3, z3), (x4, y4, z4), correspondingly indicating two points in World Line Alpha and World Line Beta. Note that a World Line is a three-dimensional line with infinite length.

Data satisfy T <= 10000, |x, y, z| <= 10,000.

 

Output

For each test case, please print two lines.

The first line contains one float number, indicating the length of best Time Tunnel.

The second line contains 6 float numbers (xa, ya, za), (xb, yb, zb), seperated by blank, correspondingly indicating the endpoints of the best Time Tunnel in World Line Alpha and World Line Beta.

All the output float number should be round to 6 digits after decimal point. Test cases guarantee the uniqueness of the best Time Tunnel.

 

Sample Input

1
1 0 1 0 1 1 0 0 0 1 1 1

 

Sample Output

0.408248
0.500000 0.500000 1.000000 0.666667 0.666667 0.666667

 

Source

2013 ACM/ICPC Asia Regional Hangzhou Online

有必要拿出空间解析几何出来看看了，尼玛用平面方程跟直线方程求解的时候除数为零了。最后没办法只能看大神的模版了。。。。

大概解题思路：先求出两直线的公垂线（两方向向量的叉积），以公垂线跟一直线形成一平面求另一直线与该平面的交点。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;
 
const double eps = 1e-;
struct Point3//三维空间点
{
    double x, y, z;
    Point3(double x=,double y=,double z=): x(x),y(y),z(z){}
    Point3 operator + (Point3 &t){return Point3(x+t.x, y+t.y, z+t.z);}
    Point3 operator - (Point3 &t) {return Point3(x-t.x, y-t.y, z-t.z);}
    Point3 operator * (double p) {return Point3(x*p, y*p, z*p);}
    Point3 operator / (double p) {return Point3(x/p, y/p, z/p);}
};
typedef Point3 Vector3;
struct Line//空间直线
{
    Point3 a,b;
};
struct Plane//空间平面
{
    Point3 a,b,c;
    Plane(){}
    Plane(Point3 a, Point3 b, Point3 c):a(a),b(b),c(c){}
};
int dcmp(double x)
{
    if(fabs(x) < eps) return ;
    return x <  ? - : ;
}
double Dot(Vector3 A,Vector3 B) { return A.x*B.x + A.y*B.y + A.z*B.z; }
double Length2(Vector3 A) { return Dot(A, A); }
Vector3 Cross(Vector3 A, Vector3 B) { return Vector3(A.y*B.z - A.z*B.y, A.z*B.x - A.x*B.z, A.x*B.y - A.y*B.x); }
 
double LineToLine(Line u,Line v,Vector3 &t)//空间直线间距离
{
    t=Cross(u.a-u.b,v.a-v.b);
    return fabs(Dot(u.a-v.a,t))/sqrt(Length2(t));
}
 
Vector3 normalVector(Plane s)//取平面法向量
{
    return Cross(s.a-s.b,s.b-s.c);
}
 
Point3 Intersection(Line l,Plane s)//空间平面与直线的交点
{
    Point3 ret = normalVector(s);
    double t = (ret.x*(s.a.x-l.a.x)+ret.y*(s.a.y-l.a.y)+ret.z*(s.a.z-l.a.z))
        /(ret.x*(l.b.x-l.a.x)+ret.y*(l.b.y-l.a.y)+ret.z*(l.b.z-l.a.z));
    ret.x = l.a.x + ( l.b.x - l.a.x ) * t;
    ret.y = l.a.y + ( l.b.y - l.a.y ) * t;
    ret.z = l.a.z + ( l.b.z - l.a.z ) * t;
    return ret;
}
 
void solve(Line A, Line B)
{
    Vector3 normal;
    double d = LineToLine(A,B,normal);
    printf("%.6lf\n",d);
    Plane pa = Plane(A.a,A.b,A.a+normal);
    Plane pb  = Plane(B.a,B.b,B.a+normal);
    Point3 u = Intersection(B,pa);
    Point3 v = Intersection(A,pb);
    printf("%.6lf %.6lf %.6lf %.6lf %.6lf %.6lf\n", v.x, v.y, v.z, u.x, u.y, u.z );
}
 
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        Line A,B;
        scanf("%lf%lf%lf", &A.a.x, &A.a.y, &A.a.z);
        scanf("%lf%lf%lf", &A.b.x, &A.b.y, &A.b.z);
        scanf("%lf%lf%lf", &B.a.x, &B.a.y, &B.a.z);
        scanf("%lf%lf%lf", &B.b.x, &B.b.y, &B.b.z);
        solve(A,B);
    }
    return ;
}