c++ - 查找点位于矩形对角线分割的矩形内的哪个扇区

Question

我有一个程序，该程序需要一个基于x和y输入返回int(0-3)的函数。返回int应该基于“扇区”，即该点位于已在其对角线上切割的矩形内。
这是我当前的代码

int liesIn(double x, double y, double w, double h){
    //x and y are relitive, so the top left corner can be thought of as the origin (0,0)
    double rectAspect = w / h;
    double pointAspect = x / y;
    if(rectAspect > pointAspect)//top of the topLeft-BottomRight line
    {
        if(y > x * rectAspect)
        {
            return 3;
        }else if(y < x * rectAspect){
            return 0;
        }
        return 4;
    }else if(rectAspect < pointAspect)
    {
        if(y > x * rectAspect)
        {
            return 2;
        }else if(y < x * rectAspect){
            return 1;
        }
        return 4;
    }else{
        return 4;//4 is the "false" condition, if the point lies on one of the 
    }
};

    std::cout << liesIn(0.25, 0.5, 1, 1) << std::endl; //should return 3, returns 3
    std::cout << liesIn(0.75, 0.1, 1, 2) << std::endl; //should return 1, returns 1
    std::cout << liesIn(0.5, 0.75, 1, 1) << std::endl; //should return 2, returns 3
    std::cout << liesIn(0.5, 0.25, 1, 1) << std::endl; //should return 0, returns 1

这几乎给出了随机结果，这是不正确的。我需要解决什么？

Answer 1

一个对角线(从0,0起)具有等式

y * w - x * h = 0

另一个对角线有等式

y * w + x * h - h * w = 0

将点x，y代入这些方程式可得出象限(结果符号告诉我们对角点位于哪一侧)。

int liesIn(double x, double y, double w, double h){

    if (y < 0 ||  y >= h || x < 0 || x >= w)
        return 5;  //outside result if needed

    if (y * w - x * h == 0 ||  y * w + x * h  - h * w  == 0)
        return 4;  //lies on diagonal 
                   //note possible issues due to float precision limitations
                   //better to compare fabs() with small epsylon value 

    int code = 0;

    if (y * w + x * h  - h * w > 0)
        code += 1;  //above second diagonal

    if (y * w - x * h > 0) {
        code += 2;    //above main diagonal
        code = 5 - code;    //flip 2/3 values to get your numbering
    }
    return code;
};

对于您的示例给出3 0 2 0-请注意，关于 (0.75, 0.1, 1, 2) << std::endl; //should return 1,的假设是错误的，0是正确的结果
和明确的例子:

 liesIn(1, 0.2, 2, 1)      0
 liesIn(1.5, 0.5, 2, 1)    1 
 liesIn(1, 0.8, 2, 1)      2
 liesIn(0.5, 0.5, 2, 1)    3
 liesIn(1, 0.5, 2, 1)      4