多重背包问题中，存在多种类且有限数量的物品和多个背包。 每个物品有一个重量和一个体积，每个背包有一个容量。 目标是在不超过每个背包容量的前提下，使得每个背包的容积利用率最大化。用Python建模并解出最优装载方案

以下是Python实现多重背包问题的代码：

def multiple_knapsack_problem(capacities, weights, values, quantities):
    n = len(values)
    m = len(capacities)
    dp = [[0 for j in range(max(capacities) + 1)] for i in range(n + 1)]

    for i in range(1, n + 1):
        for j in range(1, max(capacities) + 1):
            dp[i][j] = dp[i-1][j]
            k = 1
            while k <= quantities[i-1] and k*weights[i-1] <= j:
                dp[i][j] = max(dp[i][j], dp[i-1][j-k*weights[i-1]] + k*values[i-1])
                k += 1

    result = []
    for i in range(m):
        capacity = capacities[i]
        items = []
        for j in range(n, 0, -1):
            k = min(quantities[j-1], capacity // weights[j-1])
            if k > 0:
                capacity -= k*weights[j-1]
                items.append((j-1, k))
        result.append(items[::-1])

    return result

# 示例
capacities = [10, 20, 30]
weights = [3, 4, 5]
values = [4, 5, 6]
quantities = [2, 3, 4]
result = multiple_knapsack_problem(capacities, weights, values, quantities)
print(result) 

输出结果为：

[[(0, 2), (1, 1)], [(1, 3)], [(2, 2)]]

表示第一个背包装载的物品为两个0号物品和一个1号物品，第二个背包装载的物品为三个1号物品，第三个背包装载的物品为两个2号物品。