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 ", dist); 92 printf("%.6f %.6f %.6f %.6f %.6f %.6f ", 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 }