算法需要在一个相当优化的方式包装的矩形 [英] Algorithm needed for packing rectangles in a fairly optimal way

问题描述

Ive得到了一堆，我需要打包成尽可能小的空间（这个空间的大小应该是两个大国）。

Ive got a bunch of rectangular objects which I need to pack into the smallest space possible (the dimensions of this space should be powers of two).

我知道的各种包装算法将包中的项目，以及可能在给定的空间，但是在这种情况下，我所需要的算法来解决如何大，以致空间应该以及

I'm aware of various packing algorithms that will pack the items as well as possible into a given space, however in this case I need the algorithm to work out how large that space should be as well.

例如说，香港专业教育学院得到了下面的矩形

Eg say Ive got the following rectangles

• 128 * 32
• 128 * 64
• 在64 * 32
• 在64 * 32

它们可以被包装成一个128×128空间

They can be packed into a 128*128 space

 _________________
|128*32          |
|________________|
|128*64          |
|                |
|                |
|________________|
|64*32  |64*32   |
|_______|________|

然而，如果也有一个160 * 32和64 * 64之一，它需要一个256 * 128的空间

However if there was also a 160*32 and a 64*64 one it would need a 256*128 space

 ________________________________
|128*32          |64*64  |64*32  |
|________________|       |_______|
|128*64          |       |64*32  |
|                |_______|_______|
|                |               |
|________________|___            |
|160*32              |           |
|____________________|___________|

什么算法是有那些能够收拾一堆矩形，并确定为容器所需的大小（以2的幂，并且在每个维给定的最大大小）？

What algorithms are there that are able to pack a bunch of rectangles and determine the required size for the container (to a power of 2, and within a given maximum size for each dimension)?

推荐答案

快速和肮脏第一遍解决方案始终是一个伟大的，开始时，好像没有别的。一个比较

The quick and dirty first pass solution is always a great one to start with, as a comparison if nothing else.

从大至小贪心的位置。

把剩下的到你的打包地区最大的矩形。如果它不适合的任何地方，将其放置在扩展包区域尽可能少的地方。重复，直到你完成用最小的矩形。

Put the largest rectangle remaining into your packed area. If it can't fit anywhere, place it in a place that extends the pack region as little as possible. Repeat until you finish with the smallest rectangle.

这不是完美的，但在所有它很容易和一个不错的基础。它仍然会收拾你的原来的例子很好，给你第二同等的回答也是如此。

It's not perfect at all but it's easy and a nice baseline. It would still pack your original example perfectly, and give you an equivalent answer for the second as well.

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