题目链接

 

Description

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.

Input

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

Output

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.

Sample Input

([(]

Sample Output

()[()]

Source

 

 

题意：给了一个括号序列（只有"("  ")"  "["  "]"） 现在让添加括号，使括号序列变得匹配，要求添加最少的括号，输出这个匹配的括号序列；

 

思路：区间DP，dp[i][j]表示区间i~j匹配添加括号后区间最小长度，dp[i][j]=dp[i][k]+dp[k+1][j] ,注意当s[i]=='('&&s[j]==')' || s[i]=='['&&s[j]==']' 时，特判一下dp[i][j]=min(dp[i][j],dp[i+1][j-1]+2);  这样可以找出匹配后的序列最小长度，但是题目要求输出匹配的序列，那么可以在定义一个数组v[i][j] 标记i~j区间的断开位置，如果s[i]=='('&&s[j]==')' || s[i]=='['&&s[j]==']' 时 v[i][j]==-1, 然后在递归调用输出即可；

 

代码如下：

#include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;const int inf=0x3f3f3f3f;char s[105];int v[105][105];int dp[105][105];void print(int l,int r){    if(r
           
            dp[i][k]+dp[k+1][i+l]) { dp[i][i+l]=dp[i][k]+dp[k+1][i+l]; v[i][i+l]=k; } } if(s[i]=='('&&s[i+l]==')'||s[i]=='['&&s[i+l]==']') { if(dp[i][i+l]>dp[i+1][i+l-1]+2) { dp[i][i+l]=dp[i+1][i+l-1]+2; v[i][i+l]=-1; } } } } print(0,len-1); printf("\n"); return 0;}