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 = 20;
  const int M = 400005;
  string s[N];
  int n, minH[N], sum[N], f[1 << N], freq[N][M];
  int main() {
    cin.tie(0);
    ios_base::sync_with_stdio(0);
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      int cur = 0;
      vector<pair<int, int>> vec;
      for (int j = 0; j < s[i].size(); ++j) {
        char c = s[i][j];
        if (c == '(')
          ++cur;
        else
          --cur;
        minH[i] = min(minH[i], cur);
        vec.emplace_back(cur, -j);
      }
      sum[i] = cur;
      sort(vec.begin(), vec.end());
      int bound = s[i].size();
      for (auto [val, pos] : vec) {
        pos = -pos, val = -val;
        if (pos < bound and val >= 0) ++freq[i][val];
        bound = min(bound, pos);
      }
    }
    for (int mask = (1 << n) - 1; mask >= 0; --mask) {
      int tot = 0;
      for (int i = 0; i < n; ++i)
        if (mask & 1 << i) tot += sum[i];
      if (tot < 0) continue;
      for (int i = 0; i < n; ++i)
        if (~mask & 1 << i) {
          int h = minH[i], cur = sum[i], here = 0;
          if (tot + h < 0) {
            f[mask] = max(f[mask], freq[i][tot]);
          } else {
            f[mask] = max(f[mask], freq[i][tot] + f[mask | 1 << i]);
          }
        }
    }
    cout << f[0] << '\n';
    return 0;
  }
  