Brackets （区间DP）

个人心得：今天就做了这些区间DP，这一题开始想用最长子序列那些套路的，后面发现不满足无后效性的问题，即（，）的配对

对结果有一定的影响，后面想着就用上一题的思想就慢慢的从小一步一步递增，后面想着越来越大时很多重复，应该要进行分割，

后面想想又不对，就去看题解了，没想到就是分割，还是动手能力太差，还有思维不够。

 for(int j=;j+i<ch.size();j++)
                     {
                     if(check(j,j+i))
                         dp[j][j+i]=dp[j+][j+i-]+;
                         for(int m=j;m<=j+i;m++)
                     dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+][j+i]);
                     }

分割并一次求最大值。动态规划真的是一脸懵逼样，多思考，多瞎想吧，呼~

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

6
6
4
0
6

 #include<iostream>
 #include<cstdio>
 #include<cmath>
 #include<cstring>
 #include<iomanip>
 #include<string>
 #include<algorithm>
 using namespace std;
 int money[];
 int dp[][];
 string ch;
 const int inf=;
 int check(int i,int j){
     if((ch[i]=='('&&ch[j]==')')||(ch[i]=='['&&ch[j]==']'))
         return ;
       return ;
 }
 void init(){
    for(int i=;i<ch.size();i++)
      for(int j=;j<ch.size();j++)
         dp[i][j]=;
 }
 int main(){
     int n,m;
     while(getline(cin,ch,'\n')){
             if(ch=="end") break;
             init();
             for(int k=;k<ch.size()-;k++)
                if(check(k,k+))
                dp[k][k+]=;
                else
                 dp[k][k+]=;
                 for(int i=;i<ch.size();i++)
                {
                    for(int j=;j+i<ch.size();j++)
                     {
                     if(check(j,j+i))
                         dp[j][j+i]=dp[j+][j+i-]+;
                         for(int m=j;m<=j+i;m++)
                     dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+][j+i]);
                     }

                }
             cout<<dp[][ch.size()-]<<endl;
     }
     return ;
 }