HDU 4741 Save Labman No.004

题意：有两条异面直线，在这两条直线上分别各找一点，使得这两点间的距离最短

P1P_1是直线L1上的点，S1S_1是直线L1L_1的方向向量

P2P_2是直线L2L_2上的点，S2S_2是直线L2L_2的方向向量

这样S1×S2S_1×S_2是L1L_1和L2L_2的公垂线

不妨设P1+t1S1P_1+t_1S_1和P2+t2S2P_2+t_2S_2是满足题意的一对点，则

P1+t1S1=P2+t2S2+t(S1×S2)P_1+t_1S_1= P_2+t_2S_2+t(S_1×S_2)

移项得

t1S1=(P2−P1)+t2S2+t(S1×S2)t_1S_1= (P_2-P_1)+t_2S_2+t(S_1×S_2)

为消t2t_2，左右都叉乘S2S_2得

t1(S1×S2)=(P2−P1)×S2+t((S1×S2)×S2)t_1(S_1×S_2)= (P_2-P_1)×S_2+t((S_1×S_2)×S_2)

为了把t1的系数变成数左右都点乘S1×S2得

t1|S1×S2|2=((P2−P1)×S2)⋅(S1×S2)t_1|S_1×S_2|^2=((P_2-P_1)×S_2)\cdot(S_1×S_2)

化简得

t1=((P2−P1)×S2)⋅(S1×S2)|S1×S2|2 t_1 = \dfrac{((P_2-P_1)×S_2)\cdot(S_1×S_2)}{|S_1×S_2|^2}

同理

t2=((P1−P2)×S1)⋅(S2×S1)|S2×S1|2 t_2 = \dfrac{((P_1-P_2)×S_1)\cdot(S_2×S_1)}{|S_2×S_1|^2}

这样点就求出来了

#include <cstdio>
#include <math.h>

struct P
{
double x, y, z;
P(){}
P(double _x, double _y, double _z){x = _x; y = _y; z = _z;}
}p1, p3, p2, p4, s1, s2;
P operator + (P a, P b) {return P(a.x + b.x, a.y + b.y, a.z + b.z);}
P operator - (P a, P b) {return P(a.x - b.x, a.y - b.y, a.z - b.z);}
P operator * (double t, P a) {return P(t * a.x, t * a.y, t * a.z);}
double dot(P a, P b) {return a.x * b.x + a.y * b.y + a.z * b.z;}
P cross(P a, P b) {return P(a.y * b.z - a.z * b.y, a.z * b.x - a.x * b.z, a.x * b.y - a.y * b.x);}

int main()
{
int T;
scanf("%d", &T);

while(T--)
{
scanf("%lf%lf%lf", &p1.x, &p1.y, &p1.z);
scanf("%lf%lf%lf", &p3.x, &p3.y, &p3.z);
scanf("%lf%lf%lf", &p2.x, &p2.y, &p2.z);
scanf("%lf%lf%lf", &p4.x, &p4.y, &p4.z);
s1 = p3 - p1; s2 = p4 - p2;
double t1 = dot(cross(p2 - p1, s2), cross(s1, s2))/dot(cross(s1,s2), cross(s1,s2));
double t2 = dot(cross(p1 - p2, s1), cross(s2, s1))/dot(cross(s2,s1), cross(s2,s1));
P a = p1 + t1 * s1;
P b = p2 + t2 * s2;

printf("%.6f\n", sqrt(dot(a - b, a - b)));
printf("%.6f %.6f %.6f ", a.x, a.y, a.z);
printf("%.6f %.6f %.6f\n", b.x, b.y, b.z);
}

return 0;
}