URAL 1183. Brackets Sequence

1. 题目

http://acm.timus.ru/problem.aspx?space=1&num=1183

1183. Brackets Sequence

Time limit: 1.0 second
Memory limit: 64 MB

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 contains at most 100 brackets (characters ‘(‘, ‘)’, ‘[‘ and ‘]’) that are situated on a single line without any other characters among them.

Output

Write a single line that contains some regular brackets sequence that has the minimal possible length and contains the given sequence as a subsequence.

Sample

input output
Problem Author: Andrew Stankevich
Problem Source: 2001-2002 ACM Northeastern European Regional Programming Contest

2. 思路

给出一个由括号组成的序列,将其补充为平衡的括号序列,求最短的平衡括号序列。

区间DP。记输入为 str[0 : L-1] ,使用 int dp[i][j] 记录将 str[i : j] 补充平衡所需的最少括号个数,使用 string result[i][j] 记录将 str[i : j] 补充平衡后的最短序列,空间消耗较大,但便于输出。对于 str[i] 和 str[j] :

  1. 如果 str[i] 和 str[j] 相匹配(“()”或“[]”),则 dp[i][j] = min(dp[i + 1][j - 1], dp[i][j]) ;
  2. 然后在 str[i : j] 区间内进行DP,令 dp[i][j] = min(dp[i][k] + dp[k + 1][j], dp[i][j]),k∈[i, j) 。

在更新 dp[i][j] 的同时,要相应地更新 result[i][j] 。注意测试例包含空行, scanf("%s", str) 无法读入空行会导致WA。

3. 代码