问题描述
假设您是小偷，您入侵了一所房子.在里面发现了以下物品:
Suppose you are a thief and you invaded a house. Inside you found the following items:
一个重3磅，价值50美元的花瓶.
一个重6磅，价值30美元的银块.
这幅画重4磅，价值40美元.
镜子重5磅，价值10美元.
A vase that weights 3 pounds and is worth 50 dollars.
A silver nugget that weights 6 pounds and is worth 30 dollars.
A painting that weights 4 pounds and is worth 40 dollars.
A mirror that weights 5 pounds and is worth 10 dollars.
这个10磅大小的背包问题的解决方案是90美元.
Solution to this Knapsack problem of size 10 pounds is 90 dollars .
通过动态编程制成的表是:-
Table made from dynamic programming is :-
现在，我想知道使用该表将哪些元素放入了麻袋，然后如何回溯?
推荐答案
从您的DP表中，我们知道f [i] [w] =总重量小于1..i的子集的最大总价值.或等于w.
From your DP table we know f[i][w] = the maximum total value of a subset of items 1..i that has total weight less than or equal to w.
我们可以使用表格本身来恢复最佳包装:
We can use the table itself to restore the optimal packing:
def reconstruct(i, w): # reconstruct subset of items 1..i with weight <= w
# and value f[i][w]
if i == 0:
# base case
return {}
if f[i][w] > f[i-1][w]:
# we have to take item i
return {i} UNION reconstruct(i-1, w - weight_of_item(i))
else:
# we don't need item i
return reconstruct(i-1, w)
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