0－1背包问题(优先队列式分支限界法)python

以下是使用优先队列式分支限界法求解0-1背包问题的Python代码示例：

import queue

class Node:
    def __init__(self, level, profit, weight):
        self.level = level
        self.profit = profit
        self.weight = weight
        self.bound = 0.0

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

# 示例
n = 5
W = 10
p = [10, 5, 15, 7, 6]
w = [1, 5, 2, 4, 3]
print("最大价值为：", knapsack_bfs(n, W, p, w))

该算法使用了优先队列来实现分支限界法，能够准确地找出限定容量背包所能装载的商品的最大价值，并计算出程序运行所需要的时间。