POJ

Brackets

We give the following inductive definition of a “regular brackets” sequence:

• the empty sequence is a regular brackets sequence,
• if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
• if a and b are regular brackets sequences, then ab is a regular brackets sequence.
• no other sequence is a regular brackets sequence

For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a₁a₂ … a_n, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i₁, i₂, …, i_m where 1 ≤ i₁ < i₂ < … < i_m ≤ n, a_i1a_i2 … a_im is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input

The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output

For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input

((()))
()()()
([]])
)[)(
([][][)
end

Sample Output

6
6
4
0
6

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=105;
char s[N];
int f[N][N];
inline bool check(int i,int j){
    if(s[i]=='['&&s[j]==']') return 1;
    if(s[i]=='('&&s[j]==')') return 1;
    return 0;
}
int main(){
    while(~scanf("%s",s+1)){
        if(s[1]=='e') break;
        memset(f,0,sizeof(f));
        int n=strlen(s+1);
        for(int i=n;i>=1;i--)
            for(int j=i+1;j<=n;j++){
                if(check(i,j)) f[i][j]=f[i+1][j-1]+2;
                for(int k=i;k<=j;k++) f[i][j]=max(f[i][j],f[i][k]+f[k][j]); 
            }
        printf("%d
",f[1][n]);        
    }
}