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;
  template <typename T>
  using matrix = vector<vector<T>>;
  const long long INFL = (1LL << 62) - 1;
  const int INF = (1 << 30) - 1;
  const double EPS = 1e-7;
  const int MOD = 1e9 + 7;
  const int MAXN = 1e6 + 1;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    matrix<int> go(n);
    vector<int> open(n), close(n);
    for (int i = 0; i < n; i++) {
      string s;
      cin >> s;
      go[i].resize(s.size() + 1);
      int closect = 0, openct = 0;
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '(') {
          openct++;
        } else {
          if (openct)
            openct--;
          else
            closect++;
        }
        if (openct == 0) go[i][closect]++;
      }
      close[i] = closect;
      open[i] = openct;
    }
    int resp = 0;
    vector<int> dp(1 << n, -INF);
    dp[0] = 0;
    vector<int> state = {0};
    for (int i = 0; i < n; i++) {
      vector<int> newstate;
      for (int j = 0; j < n; j++) {
        for (int s : state) {
          if ((1 << j) & s) continue;
          int ns = s | (1 << j), total = 0;
          for (int k = 0; k < n; k++) {
            if ((1 << k) & s) total += open[k] - close[k];
          }
          if (total < 0) continue;
          if (close[j] > total) {
            resp = max(resp, dp[s] + go[j][total]);
          } else {
            dp[ns] =
                max(dp[ns], dp[s] + (total < go[j].size() ? go[j][total] : 0));
            newstate.push_back(ns);
          }
        }
      }
      sort(newstate.begin(), newstate.end());
      newstate.erase(unique(newstate.begin(), newstate.end()), newstate.end());
      state = newstate;
    }
    for (int i = 0; i < (1 << n); i++) resp = max(resp, dp[i]);
    cout << resp << '\n';
    return 0;
  }
  