1626

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 = 110;
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=0;i<n;i++){
        d[i+1][i]=0;
        d[i][i]=1;
    }
   
    for(int i=n-2;i>=0;i--){
        for(int j=i+1;j<n;j++){
            d[i][j]=maxn;
            if(match(s[i],s[j]))
                d[i][j]=min(d[i][j],d[i+1][j-1]);
            for(int k=i;k<j;k++){
                d[i][j]=min(d[i][j],d[i][k]+d[k+1][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+1][j-1]){
        printf("%c",s[i]);print(i+1,j-1);printf("%c",s[j]);
        return;
    }
    for(int k=i;k<j;k++){
        if(ans==d[i][k]+d[k+1][j]){
            print(i,k);print(k+1,j);
            return;
        }
    }
    
}
int main(int argc, const char * argv[]) {
    int T;
    scanf("%d",&T);
    getchar();
    while(T--){
        getchar();
        memset(s, 0, sizeof s);
        char ch;
        for(int i=0;(ch=getchar())!='
';i++){
            s[i]=ch;
        }
        n=strlen(s);
        
        dp();
        print(0,n-1);
        printf("
");
        if(T!=0) printf("
");
    }
    return 0;
}