UVA1626 DP经典

该博客探讨了如何解决UVA1626算法竞赛题目,这是一个关于找到最短的正规括号序列,使其包含给定字符序列作为子序列的问题。通过动态规划方法,利用状态转移方程求解,强调了理解题目描述和处理特殊情况(如空字符串)的重要性。

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Let us define a regular brackets sequence in the following way:
1. Empty sequence is a regular sequence.
2. If S is a regular sequence, then (S) and [S] are both regular sequences.
3. If A and B are regular sequences, then AB is a regular sequence.
For example, all of the following sequences of characters are regular brackets sequences:
(), [], (()), ([]), ()[], ()[()]
And all of the following character sequences are not:
(, [, ), )(, ([)], ([]
Some sequence of characters ‘(’, ‘)’, ‘[’, and ‘]’ is given. You are to find the shortest possible
regular brackets sequence, that contains the given character sequence as a subsequence. Here, a string
a1a2 … an is called a subsequence of the string b1b2 … bm, if there exist such indices 1 ≤ i1 < i2 <
… < in ≤ m, that aj = bij
for all 1 ≤ j ≤ n.
Input
The input begins with a single positive integer on a line by itself indicating the number of the cases
following, each of them as described below.
This line is followed by a blank line, and there is also a blank line between two consecutive inputs.
The input file contains at most 100 brackets (characters ‘(’, ‘)’, ‘[’ and ‘]’) that are situated on a
single line without any other characters among them.
Output
For each test case, the output must foll

### 关于UVa 307问题的动态规划解法 对于UVa 307 (Sticks),虽然通常采用深度优先搜索(DFS)+剪枝的方法来解决这个问题,但也可以尝试构建一种基于动态规划的思想去处理它。然而,在原描述中并未提及具体的动态规划解决方案[^3]。 #### 动态规划解题思路 考虑到本题的核心在于通过若干根木棍拼接成更少数量的新木棍,并使得这些新木棍尽可能接近给定的目标长度。为了应用动态规划技术,可以定义一个二维数组`dp[i][j]`表示从前i种不同类型的木棍中选取一些组合起来能否恰好组成总长度为j的情况: - 如果存在这样的组合,则`dp[i][j]=true`; - 否则`dp[i][j]=false`. 初始化时设置`dp[0][0]=true`, 表明没有任何木棍的情况下能够构成零长度。接着遍历每种类型的木棍以及所有可能达到的累积长度,更新对应的布尔值。最终检查是否存在某个k使得`sum/k * k == sum && dp[n][sum/k]`成立即可判断是否能成功分割。 这种转换方式利用了动态规划中的两个重要特性:最优化原理和重叠子问题属性。具体来说,每当考虑一根新的木棍加入现有集合时,只需要关注之前已经计算过的较短长度的结果,从而避免重复运算并提高效率[^1]. #### Python代码实现 下面给出一段Python伪代码用于说明上述逻辑: ```python def can_partition_sticks(stick_lengths, target_length): n = len(stick_lengths) # Initialize DP table with False values. dp = [[False]*(target_length + 1) for _ in range(n + 1)] # Base case initialization. dp[0][0] = True for i in range(1, n + 1): current_stick = stick_lengths[i - 1] for j in range(target_length + 1): if j >= current_stick: dp[i][j] |= dp[i - 1][j - current_stick] dp[i][j] |= dp[i - 1][j] return any(dp[-1][l] and l != 0 for l in range(target_length + 1)) # Example usage of the function defined above. stick_lengths_example = [...] # Input your data here as a list. total_sum = sum(stick_lengths_example) if total_sum % min(stick_lengths_example) == 0: result = can_partition_sticks(stick_lengths_example, int(total_sum / min(stick_lengths_example))) else: result = False print('Can partition sticks:', 'Yes' if result else 'No') ``` 需要注意的是,这段代码只是一个简化本,实际比赛中还需要进一步调整参数以适应特定输入范围的要求。此外,由于题目本身允许有多余的小段剩余未被使用,所以在设计状态转移方程时也应适当放宽条件.
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