问题描述
# 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,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],n-1),knapSack(W,n-1))
# end of function knapSack
# To test above function
val = [60,100,120]
wt = [10,20,30]
W = 50
n = len(val)
print knapSack(W,n)
为什么我们要返回最大值?我的意思是,只要我们回去,一切都很好。另一件事只是一个递归调用,如果该元素的权重大于我们的能力,那么对吗?
return val[n-1] + knapSack( W-wt[n-1],n-1)
解决方法
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