我将计算使盒子彼此适合的***方式.

I would calculate the optimal ways to fit the boxes inside each other.

• 一个Z等于一个V和一个X.
• 一个V等于两个X.
• 一个X等于两个U.
• 1个U等于5个T(对于任何合并的框，必须至少可以将2个框合并为1个框.)

此步骤在计算上可能会相当昂贵，具体取决于您有多少个盒子以及将它们彼此轻松地适配在一起.IE:如果您使用的框大小不是通用倍数，那么这将变得不那么直接和困难.请参阅 https://en.wikipedia.org/wiki/Change-making_problem 真正不错"的例子框大小的组合看起来像(您提供的示例相当不错).

This step may be rather computationally expensive depending on how many boxes you have and how easily you get fit them into each other. IE: This would be much less straightforward and harder if you had box size where they are no common multiples. See the https://en.wikipedia.org/wiki/Change-making_problem for examples of what really "nice" box size combinations would look like (the example you gave is quite nice).

将一个盒子中的所有产品移动到其他盒子中，以尽可能获得接近0的剩余空间为目标，理想的做法是先寻找会导致剩余空间恰好为0的移动(仅将所有内容放入一个盒子中并移动它放到另一个盒子里.

Move all products in one box to other boxes with the goal of get as close to 0 remaining space as you can, ideally start by looking for moves that result in exactly 0 remaining space (only by taking everything in one box and moving it to another box).

然后按照上述规则尽可能多地合并框，只要将框的数量至少减少一个即可.IE: CEF =>V (从技术上讲是 E => F ，然后是 EF => C )， A =>X，DB =>X .然后，您可以从那里进行合并. ADB = V (将 2X 框合并为一个 V 框).

Then merge the boxes as much as you can on the above rules as long as it reduces the # of boxes by at least one. IE: CEF=> V (technically it would be E => F, then EF => C), A => X, DB => X. Then you can combine it from there. ADB = V (Combine 2X boxes into a single V box).

另一个有效的选项是: DF =>X，BCE =>V，A ＝>X .在这种情况下，我们仍然将两个 X 组合为一个 V .也可能有一些解决方案，其中您可能具有 1 V 和 1 Z ，但这仅在您具有 1X 和 1V时才有意义，否则***使用 2X =>改为1V .

Another valid option is: DF => X, BCE => V, A => X. In this case, we still combine the two X into a V. There are also likely solutions where you might have 1 V and 1 Z, but that only makes sense if you had 1X and 1V, otherwise it would be better to use 2X => 1V instead.