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=0;j+i<ch.size();j++)
                    {
                    if(check(j,j+i))
                        dp[j][j+i]=dp[j+1][j+i-1]+2;
                        for(int m=j;m<=j+i;m++)
                    dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+1][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[205];
int dp[205][205];
string ch;
const int inf=999999;
int check(int i,int j){
    if((ch[i]=='('&&ch[j]==')')||(ch[i]=='['&&ch[j]==']'))
        return 1;
      return 0;
}
void init(){
   for(int i=0;i<ch.size();i++)
     for(int j=0;j<ch.size();j++)
        dp[i][j]=0;
}
int main(){
    int n,m;
    while(getline(cin,ch,'\n')){
            if(ch=="end") break;
            init();
            for(int k=0;k<ch.size()-1;k++)
               if(check(k,k+1))
               dp[k][k+1]=2;
               else
                dp[k][k+1]=0;
                for(int i=2;i<ch.size();i++)
               {
                   for(int j=0;j+i<ch.size();j++)
                    {
                    if(check(j,j+i))
                        dp[j][j+i]=dp[j+1][j+i-1]+2;
                        for(int m=j;m<=j+i;m++)
                    dp[j][j+i]=max(dp[j][j+i],dp[j][m]+dp[m+1][j+i]);
                    }

               }
            cout<<dp[0][ch.size()-1]<<endl;
    }
    return 0;
}