区间dp----括号匹配Brackets

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 sequenceFor 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 … an, 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, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim 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 Output6
6
4
0
6

#include
#include
#include
#include
using namespace std;
char s[105];
int dp[105][105];
bool check(int a,int b)
{
 if(s[a]=='(' && s[b]==')') return true;
 if(s[a]=='[' && s[b]==']') return true;
 return false;
}
int main()
{
 while(~scanf("%s",s+1))
 {
  if(s[1]=='e') break;
  memset(dp,0,sizeof(dp));
  int n=strlen(s+1);
  for(int len=2;len<=n;len++){
   for(int i=1;i+len-1<=n;i++){
    int j=i+len-1;
    for(int k=i+1;k<=j;k++){
     if(check(i,k))
      dp[i][j]=max(dp[i][j],dp[i+1][k-1]+dp[k+1][j]+2);
    }
    dp[i][j]=max(dp[i][j],dp[i+1][j]);
   }
   
  }
  printf("%d\n",dp[1][n]);
 }
 return 0;
}