Brackets

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3871		Accepted: 2028

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

 1 #include<stdio.h>
 2 #include<string.h>
 3 #define max(x,y) x>y?x:y
 4 int main(){char sequence[110];
 5     int dp[105][105],t;
 6     while(scanf("%s",sequence),strcmp(sequence,"end")){
 7         memset(dp,0,sizeof(dp));
 8         t=strlen(sequence);
 9         for(int i=t-2;i>=0;i--){
10             for(int j=i+1;j<t;j++){dp[i][j]=dp[i+1][j];
11                 for(int k=i+1;k<=j;k++){
12                     if(sequence[i]=='('&&sequence[k]==')'||sequence[i]=='['&&sequence[k]==']'){
13                         dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k][j]+2);
14                     //    printf("%d %d %c %c %d\n",i,k,sequence[i],sequence[k],dp[i][k]);
15                     }
16                 }
17             }
18         }
19         printf("%d\n",dp[0][t-1]);
20     }
21     return 0;
22 }