空间一般直线的方程是:
(x-x0)/a=(y-y0)/b=(z-z0)/c,
这是一条过(x0,y0,z0),方向矢量为{a,b,c}的直线.
假设已知点的坐标是A(e,f,g),过A点,且与{a,b,c}垂直的平面是,
a(x-e)+b(y-f)+c(z-g)=0,直线(x-x0)/a=(y-y0)/b=(z-z0)/c,与这个平面的交点是B,
再由两点的距离公式求出AB,即得.
1 import java.util.Scanner;
2
3
4 public class Main {
5
6
7
8 public static void main(String[] args) {
9
10 Scanner cin = new Scanner(System.in);
11 int n = cin.nextInt();
12
13 while(n != 0){
14 float[] enemy = new float[5];
15 for(int i = 0; i < 5; i++){
16 enemy[i] = cin.nextFloat();
17 }
18
19 float[] soldier = new float[8];
20 for(int i = 0; i < 8; i++){
21 soldier[i] = cin.nextFloat();
22 }
23
24 boolean b = hit(enemy, soldier);
25 if(b){
26 System.out.println("YES");
27 }
28 else {
29 System.out.println("NO");
30 }
31 n--;
32 }
33 cin.close();
34 }
35
36 public static boolean hit(float[] enemy, float[] soldier){
37
38 //敌人头的中心 h1,r1,x1,y1,z1
39 float r1 = enemy[1];
40 //float[] centerEnemyHead = {enemy[2], enemy[3], enemy[0] - enemy[1] + enemy[4]};
41 //警察瞄准敌人时人头的中心 h2,r2,x2,y2,z2,x3,y3,z3
42 float h2 = soldier[0];
43 float r2 = soldier[1];
44 float height = h2 * 0.9f - r2 ;//实际高度 低 0.1
45
46 float[] direction = {soldier[5], soldier[6], soldier[7]};
47
48 float a = direction[0], b = direction[1], c = direction[2];
49 float x0 = soldier[2], y0 = soldier[3], z0 = soldier[4] + height;
50 float e = enemy[2], f = enemy[3], g = enemy[0] - enemy[1] + enemy[4];
51
52 float fenmu = a*a/b + b + c*c/b;
53 float fenzi = a*a*y0/b - a*x0 + a*e + b*f + c*c*y0/b - c*z0 +c*g;
54
55 float y = fenzi / fenmu;
56
57 float x = a*(y - y0)/b + x0;
58 float z = c*(y - y0)/b + z0;
59 //两点距离平方
60 float doubleR = (x-e)*(x-e) + (y-f)*(y-f) + (z-g)*(z-g);
61
62 if(doubleR > r1*r1)
63 return false;
64 else {
65 return true;
66 }
67
68 }
69 }
