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

 //2017-05-22

 #include <cstdio>

 #include <cstring>

 #include <iostream>

 #include <algorithm>

 using namespace std;

 int dp[][];//dp[l][r]表示区间l-r中括号匹配数

 //若位置l和r匹配，dp[l][r] = max(dp[l][r], dp[l+1][r-1]+2)

 //否则，dp[l][r] = max(dp[l][r], dp[l][k]+dp[k+1][r]

 int main()

 {

     string str;

     while(cin>>str)

     {

         if(str[] == 'e')break;

         int n = str.length();

         memset(dp, , sizeof(dp));

         for(int len = ; len < n; len++){

             for(int i = ; i+len < n; i++){

                 int j = i+len;

                 if((str[i] == '(' && str[j] == ')') || (str[i] == '[' && str[j] == ']'))dp[i][j] = max(dp[i][j], dp[i+][j-]+);

                 for(int k = i; k <= j; k++)

                   dp[i][j] = max(dp[i][j], dp[i][k]+dp[k+][j]);

             }

         }

         cout<<dp[][n-]<<endl;

     }

     return ;

 }

巴特西

POJ2955(KB22-C 区间DP)