C# 0-1 已知总和和集合中零个数的背包问题
我有一个 5x5 的值表,从 0 到 3(含),所有值都是未知的。我知道每行和每列的值的总和以及零的数量。我将如何使用 C# 解决这个 0-1 背包问题并检索满足已知总和和零个数的可能解决方案?桌子总是五行五列,所以它不完全是一个传统的背包。
例如,假设我们输入:
Row[0]: Sum=4, Zeros=1
[1]: Sum=5, Zeros=1
[2]: Sum=4, Zeros=2
[3]: Sum=8, Zeros=0
[4]: Sum=3, Zeros=2
Col[0]: Sum=5, Zeros=1
[1]: Sum=3, Zeros=2
[2]: Sum=4, Zeros=2
[3]: Sum=5, Zeros=1
[4]: Sum=7, Zeros=0
我们将得到这个作为可能的解决方案:
[[ 0 1 1 1 1 ]
[ 1 0 2 1 1 ]
[ 2 1 0 0 1 ]
[ 1 1 1 2 3 ]
[ 1 0 0 1 1 ]]
在这种相当奇怪的情况下我应该采用什么类型的算法?我是否还必须编写一个类来枚举排列?
编辑澄清:问题不在于我无法列举可能性;而在于我无法列举可能性。我不知道如何有效地确定添加到任意总和的排列,同时包含指定数量的零和最多 5 个项目。
I have a 5x5 table of values from 0 to 3 inclusive with all values unknown. I know both the sum of the values and the number of zeros for each row and column. How would I go about solving this 0-1 knapsack problem using C# and retrieving the possible solutions that satisfy the known sums and number of zeros? The tables will always be five rows and five columns, so it's not quite a traditional knapsack.
For example, say we input:
Row[0]: Sum=4, Zeros=1
[1]: Sum=5, Zeros=1
[2]: Sum=4, Zeros=2
[3]: Sum=8, Zeros=0
[4]: Sum=3, Zeros=2
Col[0]: Sum=5, Zeros=1
[1]: Sum=3, Zeros=2
[2]: Sum=4, Zeros=2
[3]: Sum=5, Zeros=1
[4]: Sum=7, Zeros=0
We would get this as a possible solution:
[[ 0 1 1 1 1 ]
[ 1 0 2 1 1 ]
[ 2 1 0 0 1 ]
[ 1 1 1 2 3 ]
[ 1 0 0 1 1 ]]
What type of algorithm should I employ in this rather strange situation? Would I also have to write a class just to enumerate the permutations?
Edit for clarification: the problem isn't that I can't enumerate the possibilities; it's that I have no clue how to go about efficiently determining the permutations adding to an arbitrary sum while containing the specified number of zeros and a maximum of 5 items.
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这是代码。如果您需要任何意见,请随时询问:
Here there is the code. If you need any comment feel free to ask:
它必须有多快?我刚刚测试了一个天真的“尝试几乎所有事情”,并进行了一些早期中止,但比可能的要少,而且速度相当快(不到一毫秒)。它给出了解决方案:
如果这是您可以接受的解决方案,我可以发布代码(或者只是讨论它,它非常冗长,但基本思想很微不足道)
编辑:它也可以轻松扩展以枚举所有解决方案。它在 15 毫秒内发现了其中 400 个,并声称没有更多。这是正确的吗?
我所做的,是从 0,0 开始,尝试在该位置填写的所有值(0 尽管 min(3, rowsum[0])),填充它(从 rowsum[y] 和 colsum[x 中减去它) ] 并从 rowzero[y] 和 colzero[x] 中减去 1(如果该值为零),然后对 0,1 递归地执行此操作; 0,2; 0,3;然后在 0,4 处,我有一个特殊情况,如果剩余的 rowsum 为非负数,我只需填写它(否则,中止当前尝试 - 即在递归树中向上),以及当 y=4 时类似的情况。同时,当任何 rowsum colsum colzero 或 rowzero 变为负数时,我会中止。
当且仅当所有剩余的行和列和 colzero 和 rowzero 都为零时,当前板才是一个解决方案。所以我只是测试一下,如果是的话,将其添加到解决方案中。它不会因构造而产生任何负条目。
How fast does it have to be? I just tested a naive "try pretty much anything" with some early aborts but less than would be possible, and it was pretty fast (less than a millisecond). It gave the solution:
If that's an acceptable solution to you, I can post the code (or just discuss it, it's quite verbose but the underlying idea is trivial)
edit: it is also trivially extendable to enumerating all solutions. It found 400 of them in 15 milliseconds, and claims that there are no more than that. Is that correct?
What I did, was start at 0,0 and try all values I could fill in at that place (0 though min(3, rowsum[0])), fill it it (subtracting it from rowsum[y] and colsum[x] and subtracting one from rowzero[y] and colzero[x] if the value was zero), then recursively do this for 0,1; 0,2; 0,3; then at 0,4 I have a special case where I just fill in the remaining rowsum if it is non-negative (otherwise, abort the current try - ie go up in the recursion tree), and something similar for when y=4. In the mean time, I abort when any rowsum colsum colzero or rowzero becomes negative.
The current board is a solution if and only if all remaining rowsums columnsums colzero's and rowzero's are zero. So I just test for that, and add it to the solutions if it is one. It won't have any negative entries by construction.