我从下面的link中获取了一个代码
# A naive recursive implementation
# of 0-1 Knapsack Problem
# Returns the maximum value that
# can be put in a knapsack of
# capacity W
def knapSack(W, wt, val, n):
# Base Case
if n == 0 or W == 0:
return 0
# If weight of the nth item is
# more than Knapsack of capacity W,
# then this item cannot be included
# in the optimal solution
if (wt[n-1] > W):
return knapSack(W, wt, val, n-1)
# return the maximum of two cases:
# (1) nth item included
# (2) not included
else:
return max(
val[n-1] + knapSack(
W-wt[n-1], wt, val, n-1),
knapSack(W, wt, val, n-1))
# end of function knapSack
#Driver Code
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print knapSack(W, wt, val, n)
# This code is contributed by Nikhil Kumar Singh
我想知道最佳结果包括哪些元素。换句话说,我需要知道所取元素的索引(例如,这个示例输出为:list_of_indices = [2,1]
)
目前没有回答
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