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 N = 2e6, INF = 1e9 + 2, MOD = 1e9 + 7;
  string mas[22];
  int mn[22];
  int en[22];
  unordered_map<int, int> cnt[22];
  pair<int, int> dp[N];
  pair<int, int> add(int i, pair<int, int> res) {
    for (int j = 0; j < mas[i].size(); j++) {
      if (mas[i][j] == ')')
        res.first--;
      else
        res.first++;
      if (res.first < 0) {
        return res;
      }
      if (res.first == 0) res.second++;
    }
    return res;
  }
  void solve() {
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> mas[i];
    for (int i = 0; i < n; i++) {
      int c = 0;
      for (int j = 0; j < mas[i].size(); j++) {
        if (mas[i][j] == '(')
          c++;
        else
          c--;
        if (c < mn[i]) {
          mn[i] = c;
        }
        if (c == mn[i]) cnt[i][c]++;
      }
      en[i] = c;
    }
    for (int i = 0; i < N; i++) {
      dp[i] = {-1, -1};
    }
    dp[0] = {0, 0};
    int ans = 0;
    for (int mask = 0; mask < (1 << n); mask++) {
      if (dp[mask].first < 0) continue;
      for (int j = 0; j < n; j++) {
        if ((mask & (1 << j)) == 0) {
          int nextmask = mask ^ (1 << j);
          if (dp[mask].first >= -mn[j]) {
            pair<int, int> a = {dp[mask].first + en[j], dp[mask].second};
            if (dp[mask].first == -mn[j]) a.second += cnt[j][-dp[mask].first];
            ans = max(ans, a.second);
            if (a.first >= 0 && a.second > dp[nextmask].second) {
              dp[nextmask] = a;
            }
          } else {
            pair<int, int> a = {-1, dp[mask].second + cnt[j][-dp[mask].first]};
            ans = max(ans, a.second);
          }
        }
      }
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) solve();
  }
  