[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("
");
    return 0;
}