括号匹配

Description

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<string>
using namespace std;
int d[105][105];
int n;
string s;
void dp(){
for(int i=0;i<n;i++){d[i+1][i]=0;d[i][i]=1;}
    for(int i=n-2;i>=0;i--){
        for(int j=i+1;j<n;j++){
            d[i][j]=n;
            if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']')d[i][j]=min(d[i][j],d[i+1][j-1]);
            for(int k=i;k<j;k++){
                d[i][j]=min(d[i][j],d[i][k]+d[k+1][j]);
            }
        }
    }
}
int main(){
    while(cin>>s){
        n=s.length();
        if(s=="end")break;
        dp();
        cout<<n-d[0][n-1]<<endl;
    }
return 0;
}