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：

  1 /*
  2  * Author:  Yzcstc
  3  * Created Time:  2014/10/2 21:52:38
  4  * File Name: hdu4741.cpp
  5  */
  6 #include<cstdio>
  7 #include<iostream>
  8 #include<cstring>
  9 #include<cstdlib>
 10 #include<cmath>
 11 #include<algorithm>
 12 #include<string>
 13 #include<map>
 14 #include<set>
 15 #include<vector>
 16 #include<queue>
 17 #include<stack>
 18 #include<ctime>
 19 #define repf(i, a, b) for (int i = (a); i <= (b); ++i)
 20 #define repd(i, a, b) for (int i = (a); i >= (b); --i)
 21 #define M0(x)  memset(x, 0, sizeof(x))
 22 #define Inf  0x7fffffff
 23 #define MP make_pair
 24 #define PB push_back
 25 #define eps 1e-8
 26 #define pi acos(-1.0)
 27 using namespace std;
 28 inline int sgn(const double& x){
 29     return (x > eps) - (x < -eps);
 30 }
 31 struct point{
 32      double x, y, z;
 33      point(double _x = 0, double _y = 0, double _z = 0):x(_x), y(_y), z(_z){}
 34      void input(){
 35         scanf("%lf%lf%lf", &x, &y, &z);
 36      }
 37      double len()const{
 38          return sqrt(x * x + y * y + z * z);
 39      }
 40      point trunc(const double &l) const{
 41            double r  = l / len();
 42            return point(x * r, y * r, z * r);
 43      }
 44      bool operator!=(const point& p1)const{
 45          return !(sgn(x - p1.x) == 0 && sgn(y - p1.y) == 0 && sgn(z - p1.z) == 0);
 46      }
 47      point operator-(const point& p1)const{
 48          return point(x - p1.x, y - p1.y, z - p1.z);
 49      }
 50      point operator+(const point& p) const{
 51          return point(x + p.x, y + p.y, z + p.z);
 52      }
 53      double operator^(const point& p1)const{
 54          return x * p1.x + y * p1.y + z * p1.z;
 55      }
 56      point operator*(const point& p1)const{
 57          return point(y * p1.z - z * p1.y, z * p1.x - x * p1.z, x * p1.y - y * p1.x);
 58      }
 59      point operator*(const double& l)const{
 60          return point(x * l, y * l, z * l);
 61      }
 62 } p[101];
 63
 64
 65 int parallel(const point& u1,const point& u2,const point& v1,const point& v2){
 66     return sgn(((u1 - u2) * (v1 - v2)).len()) == 0;
 67 }
 68
 69 double line2line_dist(const point& u1, const point& u2,const point& v1,const point& v2){ //计算直线u与v距离
 70     point n =  (u1 - u2) * (v1 - v2);
 71     return fabs((u1-v1) ^ n) / n.len();
 72 }
 73
 74 point intersection(const point& u1,const point& u2, const point& v1,const point& v2){
 75     double t=((u1.x-v1.x)*(v1.y-v2.y) - (u1.y-v1.y)*(v1.x-v2.x))
 76               /((u1.x-u2.x)*(v1.y-v2.y) - (u1.y-u2.y)*(v1.x-v2.x));
 77     point p = u1 + (u2 - u1) * t;
 78     return p;
 79 }
 80
 81 void solve(){
 82      for (int i = 0; i < 4; ++i)
 83            p[i].input();
 84      point v = (p[0] - p[1]) * (p[2] - p[3]);
 85      double dist = line2line_dist(p[0], p[1], p[2], p[3]);
 86      point v2 = p[2] - p[0];
 87      if (sgn(v ^ v2) < 0) v = v * -1;
 88      v = v.trunc(dist);
 89      point p2 = intersection(v + p[0], v + p[1], p[2], p[3]);
 90      point p1 = p2 - v;
 91      printf("%.6f\n", dist);
 92      printf("%.6f %.6f %.6f %.6f %.6f %.6f\n", p1.x, p1.y, p1.z, p2.x, p2.y, p2.z);
 93
 94 }
 95
 96 int main(){
 97  //   freopen("a.in", "r", stdin);
 98  //   freopen("a.out", "w", stdout);
 99     int cas;
100     scanf("%d", &cas);
101     while (cas--){
102          solve();
103     }
104     return 0;
105 }