poj1141：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

()[()]
题解
刘汝佳的书写的很详细，我就不再解释了

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char str[105];
int f[105][105];
bool ok(int l,int r)
{
    if(str[l]=='('&&str[r]==')')return true;
    if(str[l]=='['&&str[r]==']')return true;
    return false;
}
void print(int l,int r)
{
    if(l>r)return ;
    if(l==r)
    {
        if(str[l]=='('||str[l]==')')printf("()");
        else printf("[]");
        return ;
    }
    int ans=f[l][r];
    if(ans==f[l+1][r-1]&&ok(l,r))
    {
        printf("%c",str[l]);print(l+1,r-1);printf("%c",str[r]); 
        return ;
    }
    for(int k=l ; k<r ; ++k)
        if(ans==f[l][k]+f[k+1][r])
        {
            print(l,k);
            print(k+1,r);
            return ;
        }
}
int main()
{
    gets(str);
    int len=strlen(str);
    if(!len)
    {
        printf("\n");
        return 0;
    }
    for(int i=0 ; i < len ; ++i )
        f[i][i]=1;
    for(int i=len-2 ; i>=0 ; --i)
        for(int j=i+1 ; j<len ; ++j)
        {
            f[i][j]=len;
            if(ok(i,j))f[i][j]=min(f[i][j],f[i+1][j-1]);
            for(int k=i ; k<j ; ++k)
                f[i][j]=min(f[i][j],f[i][k]+f[k+1][j]);
        }
    print(0,len-1);
    printf("\n");
    return 0;
}