Python实现01背包问题代码

当然可以，以下是Python实现01背包问题的代码：

def knapsack01(weights, values, capacity):
    n = len(weights)
    dp = [[0 for j in range(capacity+1)] for i in range(n+1)]
    for i in range(1, n+1):
        for j in range(1, capacity+1):
            if j >= weights[i-1]:
                dp[i][j] = max(dp[i-1][j], dp[i-1][j-weights[i-1]]+values[i-1])
            else:
                dp[i][j] = dp[i-1][j]
    return dp[n][capacity]

希望对你有帮助！

相关问题

Python写01背包问题代码贪心算法

好的，以下是Python实现01背包问题的贪心算法代码：

def fractional_knapsack(value, weight, capacity):
"""Return maximum value of items and their fractional amounts.

(max_value, [0/1, ...]) is returned where max_value is the maximum value of
items with total weight not more than capacity.

value[i]/weight[i] is the value per unit of weight of item i.

capacity is the maximum weight.

weight must be a list of integers.

value must be a list of integers.

capacity must be an int.
"""
index = list(range(len(value)))
# get value per unit weight in list
ratio = [v/w for v, w in zip(value, weight)]
# index is sorted according to value per unit weight in decreasing order
index.sort(key=lambda i: ratio[i], reverse=True)

max_value = 0
fractions = [0]*len(value)
for i in index:
if weight[i] <= capacity:
fractions[i] = 1
max_value += value[i]
capacity -= weight[i]
else:
fractions[i] = capacity/weight[i]
max_value += value[i]*capacity/weight[i]
break

return max_value, fractions

请问是否还有其他问题需要回答呢？

01背包问题 python代码

01背包问题是一个经典的动态规划问题。给定一组物品，每个物品有一个重量和一个价值，在限定的总重量内，如何选择物品使得总价值最大。

以下是一个Python实现的01背包问题的代码示例：

def knapsack(weights, values, capacity):
    n = len(weights)
    dp = [[0] * (capacity + 1) for _ in range(n + 1)]

    for i in range(1, n + 1):
        for w in range(capacity + 1):
            if weights[i - 1] <= w:
                dp[i][w] = max(dp[i - 1][w], dp[i - 1][w - weights[i - 1]] + values[i - 1])
            else:
                dp[i][w] = dp[i - 1][w]

    return dp[n][capacity]

# 示例使用
weights = [2, 3, 4, 5]
values = [3, 4, 5, 6]
capacity = 5
print("最大价值:", knapsack(weights, values, capacity))

在这个代码中，我们使用了二维动态规划数组`dp`，其中`dp[i][w]`表示前`i`个物品在容量为`w`的情况下的最大价值。通过遍历每个物品和每个可能的重量，更新`dp`数组，最终得到最大价值。