使用优先级队列式分支限界法解决01背包问题Python完整代码
时间: 2024-03-28 07:17:11 浏览: 13
以下是使用优先级队列式分支限界法解决01背包问题的Python完整代码:
```python
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:
bound = node.profit
j = node.level + 1
totweight = node.weight
while j < n and totweight + w[j] <= W:
totweight += w[j]
bound += p[j]
j += 1
if j < n:
bound += (W - totweight) * p[j]/w[j]
return bound
def knapsack(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.level = v.level + 1
u.weight = v.weight + w[u.level]
u.profit = v.profit + p[u.level]
if u.weight <= W and u.profit > maxprofit:
maxprofit = u.profit
u.bound = bound(u, n, W, p, w)
if u.bound > maxprofit:
pq.put(u)
u.weight = v.weight
u.profit = v.profit
u.bound = bound(u, n, W, p, w)
if u.bound > maxprofit:
pq.put(u)
return maxprofit
# Example usage
n = 4
W = 10
p = [10, 20, 30, 40]
w = [5, 4, 6, 3]
print(knapsack(n, W, p, w)) # Output: 70
```
在此示例中,我们有4个物品,每个物品都有一个对应的价值和重量。我们需要选择哪些物品放入容量为10的背包中,以最大化总价值。输出结果为70,表示我们可以选择物品1、2、4,总价值为10+20+40=70。