hdu4741

题意：给定两条异面直线，求公垂线交点及垂足。

思路:去年网络赛杭州站的题，实际上就是一个模板。。顺带用来整理补充一下计算几何模板库。

code：

/*
 * Author:  Yzcstc
 * Created Time:  2014/10/2 21:52:38
 * File Name: hdu4741.cpp
 */
#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<string>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<stack>
#include<ctime>
#define repf(i, a, b) for (int i = (a); i <= (b); ++i)
#define repd(i, a, b) for (int i = (a); i >= (b); --i)
#define M0(x)  memset(x, 0, sizeof(x))
#define Inf  0x7fffffff
#define MP make_pair
#define PB push_back
#define eps 1e-8
#define pi acos(-1.0)
using namespace std;
inline int sgn(const double& x){
    return (x > eps) - (x < -eps);
}
struct point{
     double x, y, z;
     point(double _x = 0, double _y = 0, double _z = 0):x(_x), y(_y), z(_z){}
     void input(){
        scanf("%lf%lf%lf", &x, &y, &z);
     }
     double len()const{
         return sqrt(x * x + y * y + z * z);
     }
     point trunc(const double &l) const{
           double r  = l / len();
           return point(x * r, y * r, z * r); 
     }
     bool operator!=(const point& p1)const{
         return !(sgn(x - p1.x) == 0 && sgn(y - p1.y) == 0 && sgn(z - p1.z) == 0);
     }
     point operator-(const point& p1)const{
         return point(x - p1.x, y - p1.y, z - p1.z);
     }
     point operator+(const point& p) const{
         return point(x + p.x, y + p.y, z + p.z);      
     }
     double operator^(const point& p1)const{
         return x * p1.x + y * p1.y + z * p1.z;
     }
     point operator*(const point& p1)const{
         return point(y * p1.z - z * p1.y, z * p1.x - x * p1.z, x * p1.y - y * p1.x);
     }
     point operator*(const double& l)const{
         return point(x * l, y * l, z * l);      
     }
} p[101];


int parallel(const point& u1,const point& u2,const point& v1,const point& v2){
    return sgn(((u1 - u2) * (v1 - v2)).len()) == 0;
}

double line2line_dist(const point& u1, const point& u2,const point& v1,const point& v2){ //计算直线u与v距离
    point n =  (u1 - u2) * (v1 - v2);
    return fabs((u1-v1) ^ n) / n.len();
}

point intersection(const point& u1,const point& u2, const point& v1,const point& v2){
    double t=((u1.x-v1.x)*(v1.y-v2.y) - (u1.y-v1.y)*(v1.x-v2.x)) 
              /((u1.x-u2.x)*(v1.y-v2.y) - (u1.y-u2.y)*(v1.x-v2.x));
    point p = u1 + (u2 - u1) * t;
    return p;
}

void solve(){
     for (int i = 0; i < 4; ++i)
           p[i].input();
     point v = (p[0] - p[1]) * (p[2] - p[3]);
     double dist = line2line_dist(p[0], p[1], p[2], p[3]);
     point v2 = p[2] - p[0];
     if (sgn(v ^ v2) < 0) v = v * -1;
     v = v.trunc(dist);
     point p2 = intersection(v + p[0], v + p[1], p[2], p[3]);
     point p1 = p2 - v;
     printf("%.6f
", dist);
     printf("%.6f %.6f %.6f %.6f %.6f %.6f
", p1.x, p1.y, p1.z, p2.x, p2.y, p2.z);
     
}

int main(){
 //   freopen("a.in", "r", stdin);
 //   freopen("a.out", "w", stdout);
    int cas;
    scanf("%d", &cas);
    while (cas--){
         solve();
    }
    return 0;
}