Brackets（括号最大匹配问题（区间dp））

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

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<vector>
#include<cmath>

const int maxn=1e5+5;
typedef long long ll;
using namespace std;
string str;
int dp[105][105];
int main()
{
    while(cin>>str)
    {
        if(str=="end")
        {
            break;
        }
        else
        {
            int n=str.length();
            for(int t=0;t<n;t++)
            {
                dp[t][t]=0;
            }
            for(int t=0;t<n-1;t++)
            {
                if((str[t]=='['&&str[t+1]==']')||(str[t]=='('&&str[t+1]==')'))
                {
                    dp[t][t+1]=2;
                }
                else
                {
                    dp[t][t+1]=0;
                }
            }
            for(int r=3;r<=n;r++)
            {
                for(int i=0;i<n;i++)
                {
                    int j=i+r-1;
                    if(j>n)
                    break;
                    if((str[i]=='['&&str[j]==']')||(str[i]=='('&&str[j]==')'))
                    {
                        dp[i][j]=dp[i+1][j-1]+2;
                    }
                    else 
                    dp[i][j]=0;
                    for(int k=i;k<j;k++)
                    {
                        dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]);
                    } 
                }
            }
            printf("%d
",dp[0][n-1]);
        }
    }

    return 0;
}