poj 1141 Brackets Sequence dp

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 a1 a2 ... an is called a subsequence of the string b1 b2 ... bm, if there exist such indices 1 = i1 < i2 < ... < in = m, that aj = bij for all 1 = j = n.

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

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

([(]

()[()]

题目分析：

dp要找到子问题的所在。我自己看的别人的博客才知道怎么办；dp[i][j]表示i,j之间最少需要增加的符号个数。用path记录路径。状态转移dp[i][j]=min(dp[i][k]+dp[k+1][j]);意思是从哪个点分开的话最优。当str[i]==str[j]时dp[i][j]=dp[i+1][j-1];