Java HdAcm1174

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 }