0-1背包问题有多种解法,如动态规划方法,回溯方法,分枝限界方法等。对于同一种问题,请参照教材中的算法,给出相应的程序实现。要求:(1)回溯方法;(2)分枝限界方法。
时间: 2023-07-16 17:14:05 浏览: 69
回溯方法:
```python
def backtrack_knapsack(values, weights, capacity):
n = len(values)
best_value = 0
curr_weight = 0
curr_value = 0
curr_idx = 0
taken = [0] * n
def backtrack(curr_weight, curr_value, curr_idx):
nonlocal best_value
if curr_weight > capacity:
return
if curr_value > best_value:
best_value = curr_value
if curr_idx == n:
return
taken[curr_idx] = 1
backtrack(curr_weight + weights[curr_idx], curr_value + values[curr_idx], curr_idx + 1)
taken[curr_idx] = 0
backtrack(curr_weight, curr_value, curr_idx + 1)
backtrack(curr_weight, curr_value, curr_idx)
return best_value, taken
```
分枝限界方法:
```python
import heapq
def branch_bound_knapsack(values, weights, capacity):
n = len(values)
best_value = 0
curr_weight = 0
curr_value = 0
taken = [0] * n
heap = []
heapq.heappush(heap, (-curr_value, curr_weight, curr_value, 0, taken))
while heap:
_, curr_weight, curr_value, curr_idx, taken = heapq.heappop(heap)
if curr_value > best_value:
best_value = curr_value
if curr_idx == n:
continue
# 不选当前物品
heapq.heappush(heap, (-curr_value, curr_weight, curr_value, curr_idx+1, taken.copy()))
# 选当前物品
if curr_weight + weights[curr_idx] <= capacity:
taken[curr_idx] = 1
heapq.heappush(heap, (-(curr_value+values[curr_idx]), curr_weight+weights[curr_idx], curr_value+values[curr_idx], curr_idx+1, taken.copy()))
taken[curr_idx] = 0
return best_value, taken
```