1626 - Brackets sequence——［动态规划］

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...a_n is called a subsequence of the string b₁b₂...b_m, if there exist such indices 1 ≤ i₁ < i₂ < ... < i_n ≤ m, that a_j=b_ij 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 follow the description below. The outputs of two consecutive cases will be separated by a blank line.

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

1

([(]

Sample Output

()[()]

紫书上有详解＋代码，就不废话了直接贴代码：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 using namespace std;
 const int maxn = ;
 char s[maxn];
 int d[maxn][maxn];
 int n;
 inline bool match(char a,char b){
     if((a=='('&&b==')')||(a=='['&&b==']')) return true;
     else return false;
 }
 void dp(){
     for(int i=;i<n;i++){
         d[i+][i]=;
         d[i][i]=;
     }

     for(int i=n-;i>=;i--){
         for(int j=i+;j<n;j++){
             d[i][j]=maxn;
             if(match(s[i],s[j]))
                 d[i][j]=min(d[i][j],d[i+][j-]);
             for(int k=i;k<j;k++){
                 d[i][j]=min(d[i][j],d[i][k]+d[k+][j]);
             }
         }
     }
 }
 void print(int i,int j){
     if(i>j) return;
     if(i==j){
         if(s[i]=='('||s[i]==')') printf("()");
         else printf("[]");
         return;
     }
     int ans=d[i][j];
     if(match(s[i],s[j])&&ans==d[i+][j-]){
         printf("%c",s[i]);print(i+,j-);printf("%c",s[j]);
         return;
     }
     for(int k=i;k<j;k++){
         if(ans==d[i][k]+d[k+][j]){
             print(i,k);print(k+,j);
             return;
         }
     }

 }
 int main(int argc, const char * argv[]) {
     int T;
     scanf("%d",&T);
     getchar();
     while(T--){
         getchar();
         memset(s, , sizeof s);
         char ch;
         for(int i=;(ch=getchar())!='\n';i++){
             s[i]=ch;
         }
         n=strlen(s);

         dp();
         print(,n-);
         printf("\n");
         if(T!=) printf("\n");
     }
     return ;
 }