poj 1141 Brackets Squence

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25965		Accepted: 7323		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

//注意状态转移的过程  明白什么是状态转移的本质
#include <iostream>
#include <cstdio>
#include <cstring>
#define maxn 105
using namespace std;
char s[105];
int d[105][105],len;

bool match(char a, char b)
{
    if((a=='('&&b==')')||(a=='['&&b==']'))
        return true ;
    return false ;
}

void dp()
{
      for(int i=0; i<len; i++)
      {
          d[i][i]=1;
          d[i+1][i]=0;
      }
      for(int i=len-1; i>=0; i--)
        for(int j=i+1; j<len; 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]==')')
            cout<<"()";
        else cout<<"[]";
        return;
    }


    if(match(s[i],s[j])&&d[i][j]==d[i+1][j-1])
    {
        cout<<s[i];
        print(i+1,j-1);
        cout<<s[j];
        return ;
    }
    for(int k=i; k<j; k++)
    {
        if(d[i][j]==d[i][k]+d[k+1][j])
            {
                print(i,k);
                print(k+1,j);
                return ;
            }
    }
}

int main()
{
    while(gets(s))
    {
        len=strlen(s);
        dp();
        print(0,len-1);
        cout<<endl;
    }
    return 0;
}