Brackets(区间dp）

Brackets

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3624		Accepted: 1879

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_mwhere 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

Source

Stanford Local 2004

 #include<stdio.h>
 #include<algorithm>
 #include<string.h>
 using namespace std;
 char st[] ;
 int a[][] ;
 
 bool check (int a , int b)
 {
     if (st[a] == '(' && st[b] == ')' )
         return  ;
     if (st[a] == '[' && st[b] == ']' )
         return  ;
     return  ;
 }
 
 int main ()
 {
    // freopen ("a.txt" , "r" , stdin ) ;
     while () {
         gets (st) ;
         if (strcmp (st , "end") == )
             break ;
         memset (a ,  , sizeof(a) ) ;
         int len = strlen (st) ;
         for (int o =  ; o <= len ; o++) {
             for (int i =  ;  i < len - o + ; i++) {
                 int j = i + o ;
                 for (int k = i ; k < j ; k++ ) {
                     a[i][j - ] = max (a[i][j - ] , a[i][k] + a[k + ][j - ] ) ;
                     if (check (i , j - ) ) {
                         a[i][j - ] = max (a[i][j - ] , a[i + ][j -  - ] +  ) ;
                     }
                 }
             }
         }
         printf ("%d\n" , a[][len - ] ) ;
     }
     return  ;
 }

区间dp感觉和merge sort有异曲同工之妙
从可行的最小区间出发，逐级上去，最终得到整段区间的最终结。

dp[i][j] 指[i , j]这段区间的最优解