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;
}