檢查點是否在三角形（3D）內的性能

Question

我正在尋求一種閃電般的Java方法來檢查點是否在三角形內。

我在Kasper Fauerby的論文中找到了以下c ++代碼：

typedef unsigned int uint32;
#define in(a) ((uint32&) a)
bool checkPointInTriangle(const VECTOR& point, const VECTOR& pa,const VECTOR& pb, const VECTOR& pc) {
  VECTOR e10=pb-pa;
  VECTOR e20=pc-pa;

  float a = e10.dot(e10);
  float b = e10.dot(e20);
  float c = e20.dot(e20);
  float ac_bb=(a*c)-(b*b);
  VECTOR vp(point.x-pa.x, point.y-pa.y, point.z-pa.z);

  float d = vp.dot(e10);
  float e = vp.dot(e20);
  float x = (d*c)-(e*b);
  float y = (e*a)-(d*b);
  float z = x+y-ac_bb;
  return (( in(z)& ~(in(x)|in(y)) ) & 0x80000000);
}

我想知道是否可以將此代碼段轉換為Java，如果可以，它是否會勝過我的Java代碼：

public class Util {
    public static boolean checkPointInTriangle(Vector p1, Vector p2, Vector p3, Vector point) {
        float angles = 0;

        Vector v1 = Vector.min(point, p1); v1.normalize();
        Vector v2 = Vector.min(point, p2); v2.normalize();
        Vector v3 = Vector.min(point, p3); v3.normalize();

        angles += Math.acos(Vector.dot(v1, v2));
        angles += Math.acos(Vector.dot(v2, v3));
        angles += Math.acos(Vector.dot(v3, v1));

        return (Math.abs(angles - 2*Math.PI) <= 0.005);
    }

    public static void main(String [] args) {
        Vector p1 = new Vector(4.5f, 0, 0);
        Vector p2 = new Vector(0, -9f, 0);
        Vector p3 = new Vector(0, 0, 4.5f);
        Vector point = new Vector(2, -4, 0.5f);

        System.out.println(checkPointInTriangle(p1, p2, p3, point));
    }
}

和Vector類：

public class Vector {
    public float x, y, z;

    public Vector(float x, float y, float z) {
        this.x = x; this.y = y; this.z = z;
    }

    public float length() {
        return (float) Math.sqrt(x*x + y*y + z*z);
    }

    public void normalize() {
        float l = length(); x /= l; y /= l; z /= l;
    }

    public static float dot(Vector one, Vector two) {
        return one.x*two.x + one.y*two.y + one.z*two.z;
    }

    public static Vector min(Vector one, Vector two) {
        return new Vector(one.x-two.x, one.y-two.y, one.z-two.z);
    }
}

還是有一種更快的Java方法？

提前致謝！

Answer 1

您找到的代碼（如果正確）應該比您所擁有的代碼快很多。 退貨聲明

return (( in(z)& ~(in(x)|in(y)) ) & 0x80000000);

這只是一種檢查浮點數符號位的棘手方法； 如果我沒有完全錯，那就等於：

return z < 0 && x >= 0 && y >= 0;

論文正文應證實這一點。 剩下的我想你可以改變自己。