poj1141

Brackets Sequence

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19727		Accepted: 5465		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

 

 

括号匹配DP

 

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

char str[110];
int dp[110][110];
int path[110][110];

void output(int i, int j)
{
    if(i>j)
    {
        return;
    }
    if(i==j)
    {
        if(str[i]=='('||str[i]==')')
        {
            printf("()");
        }
        else
        {
            printf("[]");
        }
    }
    else if(path[i][j]==-1)
    {
        printf("%c",str[i]);
        output(i+1,j-1);
        printf("%c",str[j]);
    }
    else
    {
        output(i,path[i][j]);
        output(path[i][j]+1,j);
    }
}

int main()
{
    int len;
    int i,j,k,r;

    while(gets(str))
    {
        len=strlen(str);
        if(len==0)
        {
            printf("\n");
            continue;
        }

        //dp
        memset(dp,0,sizeof(dp));
        for(i=0; i<len; i++)
        {
            dp[i][i]=1;
        }
        for(r=1; r<len; r++)  //r表示i,j之间的距离
        {
            for(i=0;i<len-r;i++)
            {
                j=i+r;
                dp[i][j]=65535;
                if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))
                {
                    if(dp[i][j]>dp[i+1][j-1])
                    {
                        dp[i][j]=dp[i+1][j-1];
                        path[i][j]=-1;
                    }
                }
                for(k=i;k<j;k++)
                {
                    if(dp[i][j]>(dp[i][k]+dp[k+1][j]))
                    {
                        dp[i][j]=dp[i][k]+dp[k+1][j];
                        path[i][j]=k;
                    }
                }
            }
        }
        output(0,len-1);
        printf("\n");
    }

    return 0;
}