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;
  void solve_problem() {
    const int offset = 4 * 1e5 + 10;
    int n;
    cin >> n;
    vector<vector<int>> b(n), lvp(n), m(n);
    for (int i = 0; i < n; i++) {
      string s;
      cin >> s;
      int prev_b = 0;
      b[i].resize(s.size(), 0);
      lvp[i].resize(1e6, 0);
      m[i].resize(s.size(), 0);
      m[i][0] = 2;
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '(') {
          b[i][j] = prev_b + 1;
        } else {
          b[i][j] = prev_b - 1;
        }
        prev_b = b[i][j];
        if (j != 0) {
          m[i][j] = min(m[i][j - 1], b[i][j]);
        } else {
          m[i][j] = b[i][j];
        }
        if (m[i][j] == b[i][j]) {
          ++lvp[i][b[i][j] + offset];
        }
      }
    }
    vector<pair<array<int, 2>, int>> dp(
        (1 << n), make_pair(array<int, 2>{int(-1e9), int(-1e9)}, 0));
    int max_num = 0;
    for (int mask = 1; mask < (1 << n); mask++) {
      for (int j = 0; j < n; j++) {
        if (mask & (1 << j)) {
          int sub_mask = mask ^ (1 << j);
          int cur_balance = dp[sub_mask].second;
          int cur_num = lvp[j][offset - cur_balance];
          int size = m[j].size() - 1;
          dp[mask].second = cur_balance + b[j][size];
          dp[mask].first[0] = max(dp[mask].first[0], dp[sub_mask].first[0]);
          if (cur_balance + m[j][size] < 0) {
            if (sub_mask != 0) {
              dp[mask].first[0] =
                  max(dp[mask].first[0], dp[sub_mask].first[1] + cur_num);
            } else {
              dp[mask].first[0] = max(dp[mask].first[0], cur_num);
            }
          } else {
            if (sub_mask != 0) {
              dp[mask].first[1] =
                  max(dp[mask].first[1], dp[sub_mask].first[1] + cur_num);
            } else {
              dp[mask].first[1] = max(dp[mask].first[1], cur_num);
            }
          }
          max_num = max(max_num, dp[mask].first[0]);
          max_num = max(max_num, dp[mask].first[1]);
        }
      }
    }
    cout << max(max_num, 0) << endl;
  }
  int main() {
    int t = 1;
    for (int i = 0; i < t; i++) {
      solve_problem();
    }
    return 0;
  }
  