파이썬 3.5로 작성(knapSack.py)
"""
# Returns the maximum value that can be put in a knapsack of
# capacity Wdef 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))
"""def knapSack(W, wt, val, n):
K = [[0 for x in range(W+1)] for x in range(n+1)]#Build table K[][] in bottom up manner
for i in range(n+1):
for w in range(W+1):
if i==0 or w==0:
K[i][w] == 0
elif wt[i-1] <= w:
K[i][w] = max(val[i-1] + K[i-1][w-wt[i-1]], K[i-1][w])
else:
K[i][w] = K[i-1][w]return K[n][W]
# ----main
val = [10, 40, 30, 50]
wt = [5, 4, 6, 3]
W = 10
n = len(val)
print(knapSack(W, wt, val, n))
# Time Complexity : O(wn) when w = capacity of bag, n = number of items
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