http://acm.hdu.edu.cn/showproblem.php?pid=4741
题目是输入四个点，然后确定了两条直线，然后求异面直线的距离，还有离得最近的那两个点。
纯数学题，套套数学公式就可以了。
#include "stdio.h" #include "math.h" typedef struct POINT { double x; double y; double z; }Point; Point a,b,c,d; Point m,n; double sjhls( double a1, double b1, double c1, double a2, double b2, double c2, double a3, double b3, double c3 ) { return a1*b2*c3 + a3*b1*c2 + a2*b3*c1 - a3*b2*c1 - a2*b1*c3 -a1*b3*c2; } int main() { int t; scanf( "%d", &t ); while( t-- && scanf( "%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf%lf", &a.x,&a.y,&a.z, &b.x,&b.y,&b.z, &c.x,&c.y,&c.z, &d.x,&d.y,&d.z ) == 12 ) { double m1 = a.x - b.x; double n1 = a.y - b.y; double p1 = a.z - b.z; double m2 = c.x - d.x; double n2 = c.y - d.y; double p2 = c.z - d.z; double x1 = a.x, y1 = a.y, z1 = a.z; double x2 = c.x, y2 = c.y, z2 = c.z; double t11 = sjhls( n1*p2-n2*p1, x2-x1, m2, p1*m2-p2*m1, y2-y1, n2, m1*n2-m2*n1, z2-z1, p2 ); double t22 = sjhls( n1*p2-n2*p1, m1, x2-x1, p1*m2-p2*m1, n1, y2-y1, m1*n2-m2*n1, p1, z2-z1 ); double t = sjhls( n1*p2-n2*p1, m1, m2, p1*m2-p2*m1, n1, n2, m1*n2-m2*n1, p1, p2 ); double t1 = t11/t; double t2 = -t22/t; m.x = x1 + m1 * t1; m.y = y1 + n1 * t1; m.z = z1 + p1 * t1; n.x = x2 + m2 * t2; n.y = y2 + n2 * t2; n.z = z2 + p2 * t2; double ans = sqrt( ( x2-x1+m2*t2-m1*t1 )*( x2-x1+m2*t2-m1*t1 ) + ( y2-y1+n2*t2-n1*t1 )*( y2-y1+n2*t2-n1*t1 ) + ( z2-z1+p2*t2-p1*t1 )*( z2-z1+p2*t2-p1*t1 ) ); printf( "%.6f\n", ans ); printf( "%.6f %.6f %.6f %.6f %.6f %.6f\n", m.x,m.y,m.z, n.x,n.y,n.z ); } return 0; }