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 MX = 20, MA = 400005;
  int n, mem[1 << MX], mn[MX], sum[MX], mp[MX][2 * MA];
  int dp(int mk) {
    if (__builtin_popcount(mk) == n) return 0;
    int &res = mem[mk];
    if (res != -1) return res;
    res = 0;
    int s = 0;
    for (int i = 0; i < int(n); i++)
      if (mk & (1 << i)) s += sum[i];
    for (int i = 0; i < int(n); i++)
      if (~mk & (1 << i)) {
        if (s + mn[i] < 0) {
          res = max(res, mp[i][MA - s]);
        } else {
          res = max(res, dp(mk | (1 << i)) + mp[i][MA - s]);
        }
      }
    return res;
  }
  void main_() {
    cin >> n;
    for (int i = 0; i < int(n); i++) {
      string s;
      cin >> s;
      for (char c : s) {
        if (c == '(')
          sum[i]++;
        else
          sum[i]--;
        if (sum[i] <= mn[i]) {
          mp[i][MA + sum[i]]++;
          mn[i] = sum[i];
        }
      }
    }
    memset(mem, -1, sizeof(mem));
    cout << dp(0) << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    while (t--) main_();
    return 0;
  }
  