分支限界 01背包 python
时间: 2023-11-21 11:57:02 浏览: 37
以下是分支限界法解决01背包问题的Python代码:
```python
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 = []
u = Node(-1, 0, 0)
v = Node(0, 0, 0)
maxprofit = 0
v.bound = bound(v, n, W, p, w)
q.append(v)
while len(q) > 0:
v = q.pop(0)
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:
q.append(u)
u.weight = v.weight
u.profit = v.profit
u.bound = bound(u, n, W, p, w)
if u.bound > maxprofit:
q.append(u)
return maxprofit
n = 5
W = 10
p = [10, 5, 15, 7, 6]
w = [1, 5, 2, 4, 3]
print(knapsack_bfs(n, W, p, w))
```
这段代码实现了分支限界法解决01背包问题。其中,Node类表示一个节点,包含level(当前节点所在的层数)、profit(当前节点的价值)、weight(当前节点的重量)和bound(当前节点的价值上界)。bound函数用于计算当前节点的价值上界,knapsack_bfs函数用于实现分支限界算法。