Sticks (剪枝)

本文探讨了如何使用深度优先搜索(DFS)结合剪枝技术来解决复杂问题。作者分享了两种不同的实现方式,其中一种是个人原创,另一种是从网络上获取的灵感。在第二種方法中,特别提到了在求和为0的情况下的讨论对于避免超时至关重要。
George took sticks of the same length and cut them randomly until all parts became at most 50 units long. Now he wants to return sticks to the original state, but he forgot how many sticks he had originally and how long they were originally. Please help him and design a program which computes the smallest possible original length of those sticks. All lengths expressed in units are integers greater than zero.
Input
The input contains blocks of 2 lines. The first line contains the number of sticks parts after cutting, there are at most 64 sticks. The second line contains the lengths of those parts separated by the space. The last line of the file contains zero.
Output
The output should contains the smallest possible length of original sticks, one per line.
Sample Input
POJ 1011 问题剪枝后的正解时间复杂度难以精确给出一个确定的表达式。该问题通常采用深度优先搜索结合剪枝策略求解,在未剪枝时,最坏情况下时间复杂度近似为 $O(2^n)$,因为每根木棍有放入或不放入当前大木棍两种选择[^1]。 剪枝策略能显著减少实际搜索的状态数。常见的剪枝策略包括按木棍长度降序排序、避免重复尝试相同长度的木棍、若当前木棍放入失败且当前长度为 0 或放入后达到目标长度则直接返回失败等。然而,由于剪枝的效果依赖于输入数据的分布,难以确定这些剪枝能减少多少搜索状态。 一般认为,剪枝后的时间复杂度会远小于未剪枝的指数级复杂度,但在极端情况下,剪枝可能效果不佳,时间复杂度仍可能接近指数级。在平均情况下,对于小规模输入,剪枝后的算法能较快得出结果;对于大规模输入,虽有剪枝优化,但仍可能面临性能挑战。 以下是包含部分剪枝策略的 Python 代码示例: ```python def dfs(sticks, target, current_length, start_index, used): if all(used): return True if current_length == target: return dfs(sticks, target, 0, 0, used) last_failed = -1 for i in range(start_index, len(sticks)): if not used[i] and sticks[i] != last_failed and current_length + sticks[i] <= target: used[i] = True if dfs(sticks, target, current_length + sticks[i], i + 1, used): return True used[i] = False last_failed = sticks[i] if current_length == 0 or current_length + sticks[i] == target: return False return False def solve(sticks): total_length = sum(sticks) max_length = max(sticks) sticks.sort(reverse=True) for target in range(max_length, total_length + 1): if total_length % target == 0: used = [False] * len(sticks) if dfs(sticks, target, 0, 0, used): return target sticks = [2, 2, 2, 2] print(solve(sticks)) ```
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