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;
  const int MXB = 400000 + 10;
  const int mn = 1000000000;
  signed main() {
    int n;
    cin >> n;
    vector<vector<vector<int>>> poss(n, vector<vector<int>>(MXB * 2));
    int ans = -mn;
    vector<int> bas(n);
    vector<int> dp((1 << (n + 1)), -mn);
    for (int i = 0; i < n; i++) {
      int ba = 0;
      string s;
      cin >> s;
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '(') {
          ba++;
        } else if (s[j] == ')') {
          ba--;
        }
        poss[i][MXB + ba].push_back(j);
      }
      bas[i] = ba;
    }
    dp[0] = 0;
    for (int i = 0; i < (1 << n); i++) {
      if (dp[i] != -mn) {
        for (int j = 0; j < n; j++) {
          if (!((i >> j) & 1)) {
            int tmp = 0;
            for (int ii = 0; ii < n; ii++) {
              if ((i >> ii) & 1) {
                tmp -= bas[ii];
              }
            }
            if (tmp <= 0) {
              if (poss[j][MXB + tmp - 1].size()) {
                int l = -1;
                int r = poss[j][MXB + tmp].size();
                int el = poss[j][MXB + tmp - 1][0];
                while (l + 1 != r) {
                  int mid = (l + r) / 2;
                  if (poss[j][MXB + tmp][mid] < el) {
                    l = mid;
                  } else {
                    r = mid;
                  }
                }
                ans = max(ans, dp[i] + l + 1);
              } else {
                dp[i + (1 << j)] =
                    max((long long)dp[i + (1 << j)],
                        (long long)(dp[i] + poss[j][MXB + tmp].size()));
              }
            }
          }
        }
      }
    }
    for (int i = 0; i < (1 << n); i++) {
      ans = max(ans, dp[i]);
    }
    cout << ans << "\n";
  }
  