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;
  using ll = long long;
  using ld = long double;
  template <class T>
  bool uin(T& a, const T& b) {
    return a >= b ? a = b, 1 : 0;
  }
  template <class T>
  bool uax(T& a, const T& b) {
    return a <= b ? a = b, 1 : 0;
  }
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  constexpr int inf = 1e9;
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    cin >> n;
    vector<vector<int>> a(n);
    vector<int> sum(n), low(n);
    for (auto i = (0); i < (n); ++i) {
      string s;
      cin >> s;
      a[i].assign(int((s).size()) + 1, 0);
      for (auto j = (0); j < (int((s).size())); ++j) {
        sum[i] += s[j] == '(' ? +1 : -1;
        if (uax(low[i], -sum[i])) ++a[i][low[i]];
      }
    }
    vector<int> dp(1 << n, -inf);
    dp[0] = 0;
    int ans = 0;
    for (auto S = (0); S < (1 << n); ++S) {
      uax(ans, dp[S]);
      int count = 0;
      for (auto i = (0); i < (n); ++i)
        if (S >> i & 1) count += sum[i];
      for (auto i = (0); i < (n); ++i) {
        if (S >> i & 1) continue;
        int del = 0 <= count && count < int((a[i]).size()) ? a[i][count] : 0;
        uax(count < low[i] ? ans : dp[S | 1 << i], dp[S] + del);
      }
    }
    cout << ans << '\n';
  }
  