[POJ 1141] Brackets Sequence

Brackets Sequence

 

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

()[()]

 

区间DP、重点在输出

#include<iostream>

#include<cstdio>

#include<cstring>

using namespace std;

#define max(a,b) ((a)>(b)?(a):(b))

#define INF 0x7fffffff

#define N 110



char s[N];

int ss[N][N]; //ss[i][j]=k,i到j从k位置分开添加的括号数最少 

int dp[N][N]; //dp[i][j]是在i~j区间最多括号匹配数



int judge(char c1,char c2)

{

    if(c1=='(' && c2==')') return 1;

    if(c1=='[' && c2==']') return 1;

    return 0;

}



void print(int i,int j)

{

    if(i>j) return;

    else if(i==j) 

    {

        if(s[i]=='(' || s[i]==')') cout<<"()";

        else cout<<"[]";

    }

    else

    {

        if(ss[i][j]==-1)

        {

            cout<<s[i];

            print(i+1,j-1);

            cout<<s[j];

        }

        else

        {

            print(i,ss[i][j]);

            print(ss[i][j]+1,j);

        }

    }

}

int main()

{

    int n,i,j,k,len;

    gets(s+1);

    n=strlen(s+1);

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

    {

        dp[i][i]=1;

    }

    for(len=2;len<=n;len++)

    {

        for(i=1;i<=n-len+1;i++)

        {

            j=i+len-1;

            dp[i][j]=INF;

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

            {

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

                {

                    ss[i][j]=k;

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

                }

            }

            if(judge(s[i],s[j]) && dp[i][j]>dp[i+1][j-1])

            {

                ss[i][j]=-1;

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

            }

        }

    }

    //cout<<dp[1][n]<<endl;

    print(1,n);

    printf("\n");

    return 0;

}