description: A bracket sequence is a string containing only characters "(" and ")". A regular bracket sequence (or, shortly, an RBS) is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example:
  
    * bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)"); 
    * bracket sequences ")(", "(" and ")" are not. 
  
  
  
  Let's denote the concatenation of two strings x and y as x+y. For example, "()()" + ")(" = "()())(".
  
  You are given n bracket sequences s_1, s_2, ..., s_n. You can rearrange them in any order (you can rearrange only the strings themselves, but not the characters in them).
  
  Your task is to rearrange the strings in such a way that the string s_1 + s_2 + ... + s_n has as many non-empty prefixes that are RBS as possible.
  
  Input
  
  The first line contains a single integer n (1 ≤ n ≤ 20).
  
  Then n lines follow, the i-th of them contains s_i — a bracket sequence (a string consisting of characters "(" and/or ")". All sequences s_i are non-empty, their total length does not exceed 4 ⋅ 10^5.
  
  Output
  
  Print one integer — the maximum number of non-empty prefixes that are RBS for the string s_1 + s_2 + ... + s_n, if the strings s_1, s_2, ..., s_n can be rearranged arbitrarily.
  
  Examples
  
  Input
  
  
  2
  (
  )
  
  
  Output
  
  
  1
  
  
  Input
  
  
  4
  ()()())
  (
  (
  )
  
  
  Output
  
  
  4
  
  
  Input
  
  
  1
  (())
  
  
  Output
  
  
  1
  
  
  Input
  
  
  1
  )(()
  
  
  Output
  
  
  0
  
  Note
  
  In the first example, you can concatenate the strings as follows: "(" + ")" = "()", the resulting string will have one prefix, that is an RBS: "()".
  
  In the second example, you can concatenate the strings as follows: "(" + ")" + "()()())" + "(" = "()()()())(", the resulting string will have four prefixes that are RBS: "()", "()()", "()()()", "()()()()".
  
  The third and the fourth examples contain only one string each, so the order is fixed.
solution: #include <bits/stdc++.h>
  using namespace std;
  int n, dp[2][1048576], cnt[1048576], maxn[23];
  string s[23];
  int mp[23][800003];
  int main() {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> s[i];
      int qaq = 0;
      for (int j = 0; j < s[i].size(); j++)
        if (s[i][j] == '(')
          qaq++;
        else
          qaq--, maxn[i] = min(maxn[i], qaq);
      cnt[1 << i] = qaq;
    }
    for (int i = 0; i < (1 << n); i++)
      if (__builtin_popcount(i) > 1)
        for (int j = 0; j < n; j++)
          if (i & (1 << j)) cnt[i] += cnt[1 << j];
    for (int i = 0; i < n; i++) {
      int minn = 0, qaq = 0, CNT = 0;
      for (int j = 0; j < s[i].size(); j++) {
        if (s[i][j] == '(')
          qaq++;
        else
          qaq--;
        if (qaq < minn) CNT = 0;
        minn = min(minn, qaq);
        if (qaq == minn) CNT++;
        mp[i][-minn] = CNT;
      }
    }
    for (int i = 0; i < (1 << n); i++) dp[0][i] = -19260817, dp[1][i] = -19260817;
    dp[1][0] = 0;
    for (int i = 1; i < (1 << n); i++)
      for (int j = 0; j < n; j++)
        if (i & (1 << j)) {
          dp[0][i] = max(dp[0][i], dp[0][i ^ (1 << j)]);
          if (dp[1][i ^ (1 << j)] >= 0)
            dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i] =
                max(dp[(maxn[j] + cnt[i ^ (1 << j)]) >= 0][i],
                    dp[1][i ^ (1 << j)] + mp[j][cnt[i ^ (1 << j)]]);
        }
    cout << max(dp[0][(1 << n) - 1], dp[1][(1 << n) - 1]);
  }
  