有界0-1多背包的任何伪多项式算法？

Question

假设有n个项目,例如i ₁,i ₂,...... i _n,它们中的每一个都具有已知的有界权重w ₁,w ₂,... w _n.还有一组m背包,例如k ₁,k ₂和k _m.背包是同质的,它们都具有相同的容量W.功能F可以确定每个背包的得分.F的输入是每个背包中的项目.所以,

Score of each knapsack i = F(Items in knapsack i)

现在我想把一些物品放在背包里,这样:

1. 背包中物品的重量不超过其容量W.
2. 所有背包的得分总和最大

是否存在针对此问题的多项式时间解决方案？

注意:问题是0-1,即每个项目都可以选择.所有问题参数都是有界的.

编辑1:是不是可以减少这个问题来装箱,然后得出结论,这是一个NP难问题？

编辑2在此问题中,每个项目都有三个属性,例如属性a _i,b _i和c _i.F函数是一个线性函数,它获取其中项目的属性并生成输出.

编辑3:本文似乎为多背包问题提出了一个精确的解决方案.可以用在我的情况下吗？

Answer 1

这个怎么样？

给定 Haskell 中 0-1 背包问题的标准动态解，可在此处找到，

inv = [("map",9,150), ("compass",13,35), ("water",153,200), ("sandwich",50,160),
       ("glucose",15,60), ("tin",68,45), ("banana",27,60), ("apple",39,40),
       ("cheese",23,30), ("beer",52,10), ("cream",11,70), ("camera",32,30),
       ("tshirt",24,15), ("trousers",48,10), ("umbrella",73,40), 
       ("trousers",42,70), ("overclothes",43,75), ("notecase",22,80),
       ("sunglasses",7,20), ("towel",18,12), ("socks",4,50), ("book",30,10)]

knapsack = foldr addItem (repeat (0,[])) where
    addItem (name,w,v) list = left ++ zipWith max right newlist where
        newlist = map (\(val, names)->(val + v, name:names)) list
        (left,right) = splitAt w list

main = print $ (knapsack inv) !! 400

我们添加了一个填充机制，将库存排列顺序放置在下一个有空间的背包中，

stuff (name,w,v) left (v2,[]) = (v2,left)
stuff (name,w,v) left (v2,(cap, lst):xs) =
  if w <= cap 
     then (v + v2, left ++ [(cap - w, (name,w,v):lst)] ++ xs) 
     else stuff (name,w,v) (left ++ [(cap,lst)]) (v2,xs)

并将其替换为映射函数。把它们放在一起：

inv = [("map",9,150), ("compass",13,35), ("water",153,200), ("sandwich",50,160),
       ("glucose",15,60), ("tin",68,45), ("banana",27,60), ("apple",39,40),
       ("cheese",23,30), ("beer",52,10), ("cream",11,70), ("camera",32,30),
       ("tshirt",24,15), ("trousers",48,10), ("umbrella",73,40), 
       ("trousers",42,70), ("overclothes",43,75), ("notecase",22,80),
       ("sunglasses",7,20), ("towel",18,12), ("socks",4,50), ("book",30,10)]

capacity = 200::Int
numKnapsacks = 3

stuff (name,w,v) left (v2,[]) = (v2,left)
stuff (name,w,v) left (v2,(cap, lst):xs) =
  if w <= cap 
     then (v + v2, left ++ [(cap - w, (name,w,v):lst)] ++ xs) 
     else stuff (name,w,v) (left ++ [(cap,lst)]) (v2,xs)

knapsack = foldr addItem (repeat (0, replicate numKnapsacks (capacity,[]))) 
  where addItem (name,w,v) list = left ++ zipWith max right newlist 
          where newlist = map (stuff (name,w,v) []) list
                (left,right) = splitAt w list

main = print $ (knapsack inv) !! 600

输出（总价值以及每个背包的剩余重量和内容）：

*Main> main
(1062,[(1,[("map",9,150),("tshirt",24,15),("trousers",42,70),
           ("overclothes",43,75),("notecase",22,80),("sunglasses",7,20),
           ("towel",18,12),("socks",4,50),("book",30,10)]),
       (0,[("compass",13,35),("cheese",23,30),("cream",11,70),
           ("camera",32,30),("trousers",48,10),("umbrella",73,40)]),
       (1,[("sandwich",50,160),("glucose",15,60),("tin",68,45),("banana",27,60),
           ("apple",39,40)])])