Brackets

Time Limit: 1000MSMemory Limit: 65536K

Total Submissions: 7823Accepted: 4151

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

Source

Stanford Local 2004

 1 //2017-05-22
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <iostream>
 5 #include <algorithm>
 6 
 7 using namespace std;
 8 
 9 int dp[110][110];//dp[l][r]表示区间l-r中括号匹配数
10 //若位置l和r匹配，dp[l][r] = max(dp[l][r], dp[l+1][r-1]+2)
11 //否则，dp[l][r] = max(dp[l][r], dp[l][k]+dp[k+1][r]
12 
13 int main()
14 {
15     string str;
16     while(cin>>str) 
17     {
18         if(str[0] == 'e')break;
19         int n = str.length();
20         memset(dp, 0, sizeof(dp));
21         for(int len = 1; len < n; len++){
22             for(int i = 0; i+len < n; i++){
23                 int j = i+len;
24                 if((str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']'))dp[i][j] = max(dp[i][j], dp[i+1][j-1]+2);
25                 for(int k = i; k <= j; k++)
26                   dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+1][j]);
27             }
28         }
29         cout<<dp[0][n-1]<<endl;
30     }
31 
32     return 0;
33 }

POJ2955(KB22-C 区间DP)