用递归算法求解简单装载问题代码

递归算法求解简单装载问题通常是指解决“0-1背包”问题，这是一个经典的动态规划问题，也可以通过分治法（ Divide and Conquer）转化为递归。假设有一个背包可以承载一定重量W，有n件物品，每件物品有自己的重量w[i]和价值v[i]。目标是找到一种组合方式，使得背包内的物品总重量不超过W，同时总价值最大。

以下是使用Python编写的递归求解0-1背包问题的基本代码：

def knapsack_recursive(W, wt, val, n):
    # Base cases
    if n == 0 or W == 0:
        return 0

    # If weight of the nth item is more than Knapsack capacity
    # then this item cannot be included in the optimal solution
    if wt[n - 1] > W:
        return knapsack_recursive(W, wt, val, n - 1)

    # Include the nth item, recur for remaining capacity
    # and weight without the nth item
    else:
        include = val[n - 1] + knapsack_recursive(W - wt[n - 1], wt, val, n - 1)
        
        # Exclude the nth item, recur for remaining capacity
        exclude = knapsack_recursive(W, wt, val, n - 1)
        
        # Return the maximum of two possibilities
        return max(include, exclude)

# 示例
val = [60, 100, 120]
wt = [10, 20, 30]
W = 50
n = len(val)
print("最大价值:", knapsack_recursive(W, wt, val, n))