写出分支限界法的01背包问题的具体代码

以下是使用分支限界法解决01背包问题的 Python 代码：

from queue import PriorityQueue

class Node:
    def __init__(self, level, profit, weight, bound):
        self.level = level
        self.profit = profit
        self.weight = weight
        self.bound = bound

    def __lt__(self, other):
        return self.bound > other.bound

def bound(node, n, W, p, w):
    if node.weight >= W:
        return 0
    else:
        profit_bound = node.profit
        j = node.level + 1
        totweight = node.weight
        while (j < n) and (totweight + w[j] <= W):
            totweight += w[j]
            profit_bound += p[j]
            j += 1
        if j < n:
            profit_bound += (W - totweight) * p[j] / w[j]
        return profit_bound

def knapsack_bnb(n, W, p, w):
    pq = PriorityQueue()
    v = Node(-1, 0, 0, 0)
    u = Node(-1, 0, 0, bound(v, n, W, p, w))
    maxprofit = 0
    pq.put(u)
    while not pq.empty():
        v = pq.get()
        if v.bound > maxprofit:
            u = Node(v.level + 1,
                     v.profit + p[v.level + 1],
                     v.weight + w[v.level + 1],
                     bound(v, n, W, p, w))
            if u.weight <= W and u.profit > maxprofit:
                maxprofit = u.profit
            if u.bound > maxprofit:
                pq.put(u)
            u = Node(v.level + 1,
                     v.profit, v.weight, bound(v, n, W, p, w))
            if u.bound > maxprofit:
                pq.put(u)
    return maxprofit

其中，`Node` 类表示决策树中的一个节点，`bound` 函数计算当前节点的上界，`knapsack_bnb` 函数实现了分支限界法求解 01 背包问题。输入参数 `n` 表示物品的个数，`W` 表示背包的容量，`p` 表示每个物品的价值，`w` 表示每个物品的重量。输出结果为背包所能容纳的最大价值。