分治限界FIFO队列01背包问题伪代码
时间: 2023-10-15 17:02:42 浏览: 96
以下是分治限界FIFO队列01背包问题的伪代码:
```
function divide_and_conquer_knapsack(capacity, items):
queue = FIFO queue
queue.enqueue(items)
best_value = 0
while (not queue.empty()):
current_items = queue.dequeue()
# Calculate the value and weight of the current items
current_value = 0
current_weight = 0
for item in current_items:
current_value += item.value
current_weight += item.weight
# If the current items are better than the previous best, update the best value
if current_weight <= capacity and current_value > best_value:
best_value = current_value
# If the current items can be split further, add the new subproblems to the queue
if len(current_items) > 1:
left_items = current_items[:len(current_items) // 2]
right_items = current_items[len(current_items) // 2:]
left_value = 0
left_weight = 0
right_value = 0
right_weight = 0
for item in left_items:
left_value += item.value
left_weight += item.weight
for item in right_items:
right_value += item.value
right_weight += item.weight
left_bound = left_value + (capacity - left_weight) * (right_value / right_weight) if right_weight > 0 else 0
right_bound = right_value + (capacity - right_weight) * (left_value / left_weight) if left_weight > 0 else 0
# If the left subproblem has a better bound, add it to the queue
if left_bound > best_value:
queue.enqueue(left_items)
# If the right subproblem has a better bound, add it to the queue
if right_bound > best_value:
queue.enqueue(right_items)
# Return the best value found
return best_value
```
在这个算法中,我们使用了一个FIFO队列来存储待处理的子问题。我们首先将所有物品作为一个子问题加入队列中。然后,我们反复从队列中取出一个子问题,并且计算它的价值和重量。如果该子问题的重量不超过背包容量,并且价值比之前得到的最优解更好,我们就更新最优解。如果该子问题可以进一步分解,我们就根据限界函数的结果将其拆分成两个子问题,并计算每个子问题的限界。如果左子问题的限界比当前最优解更好,我们就将其加入队列中。同样,如果右子问题的限界比当前最优解更好,我们也将其加入队列中。最终,我们返回找到的最优解。
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