POJ2955――Brackets - c++编程基础

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3341		Accepted: 1717

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, andif 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 ofs. That is, you wish to find the largest m such that for indicesi₁, 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

也是一道十分经典的区间dp题，我们用dp[i][j]表示从i到j，最大括号匹配数
如果第i个括号无法在[i+1, j]中匹配，那么dp[i][j] = dp[i+1][j]；
否则如果在区间[i,j]中找到一个k，使得i和k配对，那么区间就被划分为2段，[i+1, k - 1]和[k+1,j]

所以dp[i][j] = max(dp[i +1][j], dp[i + 1][k - 1] + dp[k +1][j] + 2)

#include   
#include 
   
     #include 
    
      #include 
     
       #include 
      
        #include 
       
         #include 
        
          #include 
         
           #include 
          
            #include 
           
             #include 
            
              using namespace std; char str[110]; int dp[110][110]; int main() { while (~scanf("%s", str), str[0] != 'e') { int len = strlen(str); memset (dp, 0, sizeof(dp)); for (int i = len - 1; i >= 0; --i) { for (int j = i + 1; j < len; ++j) { dp[i][j] = dp[i + 1][j]; for (int k = i + 1; k <= j; ++k) { if ((str[i] == '(' && str[k] == ')') || (str[i] == '[' && str[k] == ']')) { dp[i][j] = max(dp[i][j], dp[i + 1][k - 1] + dp[k + 1][j] + 2); } } } } printf("%d\n", dp[0][len - 1]); } return 0; }