从网格中的一个字段到另一个字段的正交步数？
如何计算网格中从一个字段到另一个字段正交移动的“步数”？
我正在为我正在开发的游戏实现一个 A* 寻路系统，但这个简单的数学运算阻碍了我。
我可能应该重新上三年级。哈哈。
How do I calculate the amount of "steps" there is from one field to another in a grid, moving orthogonally?
I am implementing an A* pathfinding system for a game that I am developing, and this simple mathematical operation is in my way.
I should probably re-attend third grade. Haha.
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如果我理解正确的话，我认为你只需将必要的 x,y 运动相加即可。给定两个点
(x1,y1)和(x2,y2)，那么距离（假设“正交移动”意味着仅水平和/或垂直移动）则为：例如，从位置（1,1）移动到（3,4）意味着向右移动2个空格，向上移动3个空格，总共5个。
abs(1-3)+abs(1-4 ) = 2 + 3 = 5If I understand correctly, I think you just add up the x,y movements necessary. Given two points
(x1,y1)and(x2,y2), then the distance (assuming "moving orthogonally" means moving only horizontally and/or vertically) then it is:For example, moving from position (1,1) to (3,4) means moving 2 spaces to the right and 3 spaces up for a total of 5.
abs(1-3)+abs(1-4) = 2 + 3 = 5我确实相信这是一个简单的数学问题。
当然，您知道起始 x / y 值和结束 x / y 值。要获得两者之间的距离，请执行以下操作：
dist = sqrt(dx^2 + dy^2 )
其中 dx 是点的 x 坐标之间的差
其中 dy 是点的 y 坐标之间的差。
所以，举例来说。假设坐标 A 是 A(15,20)，坐标 B 是 B(35,5)；
dx = 35 - 15 = 20；
dy = 20-5 = 15;
所以;
AB 之间的距离 = sqrt(20^2 + 15^2)
= 25.0 单位。
现在，对于您的最终答案，这取决于您的程序中的“步骤”有多少个单元。如果步长为 5 个单位 (25/5)，则从 A 点到 B 需要 5 步。
I do believe this is a matter of simple math.
Surely, you know your starting x / y values, and your ending x / y values. To get the distance between the two, you do this:
dist = sqrt(dx^2 + dy^2 )
Where dx is the difference between the x-coordinates of the points
Where dy is the difference between y-coordinates of the points.
So, for example. Lets say co-ordinate A is A(15,20) and co-ordinate B is B(35,5);
dx = 35 - 15 = 20;
dy = 20-5 = 15;
Therefore;
dist between AB = sqrt(20^2 + 15^2)
= 25.0 units.
Now for your final answer, this depends how many units a "step" is in your program. If a step is 5 units, (25/5) than there is 5 steps needed to get from point A to B.