POJ 1141 Brackets Sequence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19636		Accepted: 5437		Special Judge

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

Northeastern Europe 2001

//终于找到黑书上的第一个动规了

//本来1Y的、郁闷的是我乐极生悲了，在空串的时候我不输空我居然输了个0出来、、、

//昨晚找到的这题，升级版吧、黑书上只是求出最少需要添加几个，而这里需要打印出来

//我利用st数组记录状态的转移，写的有点复杂、哈哈，明天就回家

//今天的打算就是把这题A掉、呵呵、回家休养生息几天、

#include <iostream>

#include <stdio.h>

#include <string.h>

#include <algorithm>

#include <queue>

#include <cmath>

#include <vector>

using namespace std;

struct node

{

    bool flag;

    int i1,j1,i2,j2;

};

node st[102][102];

int dp[122][122];

bool b[122];

char s[122];

int l;

void dfs (node p)

{

    if(p.flag==0)

    {

        if(p.i1==p.j1)

        {

            b[p.i1]=1;

            return;

        }

        else

        {

          if(p.i1>p.j1) return;

        }

        dfs(st[p.i1][p.j1]);

    }

    else

    {

        dfs(st[p.i1][p.j1]);

        dfs(st[p.i2][p.j2]);

    }

}

void del()

{

    int i,k,j,t;

    i=0;

    dp[0][0]=1;

    st[i][i].flag=0;

    st[i][i].i1=st[i][i].j1=i;



    for(i=1;i<l;i++)

     {

         dp[i][i]=1,dp[i][i-1]=0;

         st[i][i].flag=0;

         st[i][i].i1=st[i][i].j1=i;

         st[i][i-1].flag=0;

         st[i][i-1].i1=i;

         st[i][i-1].j1=i-1;



     }

    for(k=2;k<=l;k++)

    {

        for(i=0;i+k<=l;i++)

        {

            j=i+k-1;

            dp[i][j]=1000;

            if((s[i]=='('&&s[j]==')')||(s[i]=='['&&s[j]==']'))

              {

                  dp[i][j]=dp[i+1][j-1];

                  st[i][j].flag=0;

                  st[i][j].i1=i+1;

                  st[i][j].j1=j-1;

              }

          for(t=i;t<j;t++)

            {

                if(dp[i][j]>dp[i][t]+dp[t+1][j])

                {

                   dp[i][j]=dp[i][t]+dp[t+1][j];

                   st[i][j].flag=1;

                   st[i][j].i1=i;

                   st[i][j].j1=t;

                   st[i][j].i2=t+1;

                   st[i][j].j2=j;

                }

            }

        }

    }

    memset(b,0,sizeof(b));

    dfs(st[0][l-1]);

    for(i=0;i<l;i++)

    if(b[i])

    {

      if(s[i]=='('||s[i]==')')

       printf("%c%c",'(',')');

       else

       printf("%c%c",'[',']');

    }

    else

     printf("%c",s[i]);

    printf("\n");

}

int main()

{



    while(gets(s)!=NULL)

    {

        if(l=strlen(s))

           del();

        else

          printf("\n");



    }

    return 0;

}