Brackets（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
题目大意：
给定一个字符串,求最长合法子序列的长度。合法子序列就是每对括号一一匹配，如(), [], (()), ()[], ()[()]等等。
这题可以逆向想，求需要加多少括号使它合法，然后总长度减去它就行了。

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
int dp[205][205];
int main()
{
    ios::sync_with_stdio(false);
    string s;
    while(cin>>s)
    {
        if(s=="end") break;
        memset(dp,0,sizeof dp);
        int n=s.size();
        for(int i=0;i<n;i++)
            dp[i][i]=1;
        for(int l=1;l<n;l++)
            for(int i=0;i<n;i++)
            {
                int j=i+l;
                dp[i][j]=1<<30;
                if(s[i]=='('&&s[j]==')'||s[i]=='['&&s[j]==']')
                    dp[i][j]=dp[i+1][j-1];
                for(int k=i;k<j;k++)
                    dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]);
            }
        cout<<n-dp[0][n-1]<<'
';
    }
    return 0;
}