矩形项目的优化网格 [英] optimized grid for rectangular items

问题描述

我有N个长宽比为Aitem(X:Y)的矩形项目.
我有一个长宽比为Aview的矩形显示区域

I have N rectangular items with an aspect ratio Aitem (X:Y).
I have a rectangular display area with an aspect ratio Aview

项目应以表格形式排列(即r行，c列).

The items should be arranged in a table-like layout (i.e. r rows, c columns).

什么是理想的网格行x列，以便单个项目最大? (当然，行*列> = N-即可能有未使用的"网格位置).

what is the ideal grid rows x columns, so that individual items are largest? (rows * colums >= N, of course - i.e. there may be "unused" grid places).

一种简单的算法可以遍历行= 1..N，计算所需的列数，并使行/列对具有最大的项.

A simple algorithm could iterate over rows = 1..N, calculate the required number of columns, and keep the row/column pair with the largest items.

我想知道是否有一个非迭代算法(例如，对于Aitem = Aview = 1，行/列可以由sqrt(N)近似).

I wonder if there's a non-iterative algorithm, though (e.g. for Aitem = Aview = 1, rows / cols can be approximated by sqrt(N)).

推荐答案

注意:我不太理解Frédéric的回答，所以我自己解决了这个问题，并提出了似乎相同的解决方案.我想我也最好解释一下我做了些什么，以防它有所帮助.

Note: I couldn't quite understand Frédéric's answer, so I worked the problem out myself and came up with what appears to be the same solution. I figured I might as well explain what I did in case it is helpful.

首先，我将视图的长宽比标准化为项目的长宽比. (我假设您不想旋转这些项目.)

First I normalized the aspect ratio of the view to that of the items. (I'm assuming you don't want to rotate the items.)

a = (view_width/view_height) / (item_width/item_height)

现在将宽度/高度比a的矩形与正方形打包在一起，等效于将视图与项目打包在一起.理想的情况是让我们的网格(现在为正方形)完全填充矩形，这将给我们

Now packing a rectangle of width/height ratio a with squares is equivalent to packing the view with items. The ideal case would be for our grid (of squares now) to fill the rectangle completely, which would give us

a = c/r

其中r和c是行数和列数:

N = r*c

将这两个方程相乘/除就得出

Multiplying/dividing these two equations gives us

N*a = c^2              N/a = r^2
c = sqrt(N*a)          r = sqrt(N/a)

如果网格是完美的，则r和c将是整数，但如果不是，则必须尝试Frédéric提到的三个选项，并保持r*c最小但仍大于N的选项. :

If the grid is perfect, r and c will be integers, but if not, you have to try the three options Frédéric mentioned and keep the one where r*c is smallest but still more than N:

• floor(r), ceil(c)
• ceil(r), floor(c)
• ceil(r), ceil(c)

• floor(r), ceil(c)
• ceil(r), floor(c)
• ceil(r), ceil(c)

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