HDU 4741 Save Labman No.004 计算几何 数学

题目链接

这道题会暴露很多孩纸的高数学的好不好。如何计算三维直线的最短距离和垂足的坐标，学习链接

之后的就是代入公式计算了。

 1 #include <stdio.h>
 2 #include <math.h>
 3 struct Point {
 4     double x, y, z;
 5     Point(double x = 0, double y = 0, double z = 0):x(x), y(y), z(z) {}
 6     void read() {
 7         scanf("%lf%lf%lf", &x, &y, &z);
 8     }
 9     void write() {
10         printf("%.6lf%.6lf%.6lf", x, y, z);
11     }
12     Point operator + (const Point &b) {
13         return Point(x + b.x, y + b.y, z + b.z);
14     }
15     Point operator - (const Point &b) {
16         return Point(x - b.x, y - b.y, z - b.z);
17     }
18     Point operator * (double k) {
19         return Point(x * k, y * k, z * k);
20     }
21     Point operator / (double k) {
22         return Point(x / k, y / k, z / k);
23     }
24     Point cross(const Point &b) {
25         return Point(y * b.z - z * b.y, z * b.x - x * b.z, x * b.y - y * b.x);
26     }
27     double len() {
28         return sqrt(x * x + y * y + z * z);
29     }
30     double operator * (const Point &b) {
31         return x * b.x + y * b.y + z * b.z;
32     }
33 };
34 typedef Point Vector;
35 int main() {
36     int T;
37     scanf("%d", &T);
38     while (T--) {
39         Point A, B, C, D;
40         A.read();
41         B.read();
42         C.read();
43         D.read();
44         Vector AB= B - A, CD = D - C;
45         Vector n = AB.cross(CD), AC = C - A, CA = A - C;
46         double nlen = n.len(), dis = n * AC / nlen;
47         double t1 = AC.cross(CD) * AB.cross(CD) / nlen / nlen;
48         double t2 = CA.cross(AB) * CD.cross(AB) / nlen / nlen;
49         Point E = A + (AB * t1);
50         Point F = C + (CD * t2);
51         if (dis < 0)
52             dis = -dis;
53         printf("%.6lf\n", dis);
54         E.write();
55         putchar(' ');
56         F.write();
57         putchar('\n');
58     }
59     return 0;
60 }