2D等距引擎-数学问题-多维数据集选择-菱形形状图 [英] 2D isometric engine - Math problems - Cube selection - diamond shape map

问题描述

当我创建一个包含cube对象(其中是我的一类)的std::vector层时，我会计算坐标:

I calculate my coordinates when i create a layer with a std::vector, filled with cube objects(wich is a class of mine):

for(int J = 0; J < mapSize; J++)
{
    for(int I = 0; I < mapSize; I++)
    {
        x = (J - I) * (cubeSize/2);
        y = (J + I) * (cubeSize/4);

        c = new cube(cubeSize, x, y, z, I, J);
        cs.push_back(*c);
    }
}

我想这样做:cs[getCubeByID(mouseX, mouseY)].setTexture(...);

使用示例:I-J [0, 0]中的多维数据集在cubes数组中具有number 0.如果我click on 0,0我知道这个电话号码.

Example of use: The cube in I-J [0, 0] have the number 0 in the cubes array. if i click on 0,0 i got this number.

我们给了我公式，让J或I在注释中带有一对x，y，非常感谢. 我只需要像我给的例子一样将这对I-J转换为数组的条目号即可.

We gave me the formula to get a J or a I with a pair of x,y in the comments, thanks a lot. I only need to convert this pair of I-J to the entry number of my array like the example i gave.

我尝试过:int entry = (J - 1) * size + (I - 1);并且所选的多维数据集与我想要的多维数据集相距不远，但仍不是正确的公式.模块化算术可以解决我的问题，但是我不知道它是如何工作的.

I tried : int entry = (J - 1) * size + (I - 1); and the selected cube is not so far from the one i want but still not the right formula. Modular arithmetic can fix my problem but i don't understand how it's working.

推荐答案

所以你有

x = (J - I) * (cubeSize/2);
y = (J + I) * (cubeSize/4);

，并且您想据此计算I和J(以及因此索引为I + J*mapSize)，对吗?这是一个由两个方程组成的线性系统.

and you want to compute I and J (and therefore the index which is I + J*mapSize) from that, right? It's a linear system of two equations.

J - I = x * 2 / cubeSize
J + I = y * 4 / cubeSize

I = (y * 2 - x) / cubeSize
J = (y * 2 + x) / cubeSize

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