0-1背包问题的Python程序

# 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
  
# To test above function
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

输出：

# A Dynamic Programming based Python 
# Program for 0-1 Knapsack problem
# Returns the maximum value that can 
# be put in a knapsack of capacity W
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]
  
# Driver program to test above function
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print(knapSack(W, wt, val, n))
  
# This code is contributed by Bhavya Jain

输出：

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