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;
  auto fio = (ios::sync_with_stdio(0), cin.tie(0), cout.tie(0));
  int32_t main() {
    int n;
    cin >> n;
    vector<string> s(n);
    for (int i = 0; i < n; i++) cin >> s[i];
    vector<vector<int>> pfx(n), cnt(n);
    vector<int> all(n);
    for (int i = 0; i < n; i++) {
      int sz = s[i].size();
      int SH = sz + 2;
      cnt[i].resize(2 * sz + 4);
      pfx[i].resize(2 * sz + 4, -1);
      int bal = 0;
      for (int j = 0; j < sz; j++) {
        bal += s[i][j] == '(' ? +1 : -1;
        if (pfx[i][bal + SH] == -1) {
          pfx[i][bal + SH] = cnt[i][bal + 1 + SH];
        }
        cnt[i][bal + SH]++;
      }
      all[i] = bal;
    }
    auto smax = [](int& a, int b) { return a = max(a, b); };
    int ans = 0;
    vector<int> dp(1 << n, -1);
    dp[0] = 0;
    for (int msk = 0; msk < 1 << n; msk++) {
      if (dp[msk] == -1) continue;
      int bal = 0;
      for (int i = 0; i < n; i++)
        if (msk >> i & 1) bal += all[i];
      if (bal < 0) continue;
      for (int i = 0; i < n; i++) {
        int sz = s[i].size();
        int SH = sz + 2;
        if (~msk >> i & 1) {
          if (bal > sz)
            smax(dp[msk | 1 << i], dp[msk]);
          else {
            if (pfx[i][-bal - 1 + SH] != -1) {
              smax(ans, dp[msk] + pfx[i][-bal - 1 + SH]);
            } else {
              smax(dp[msk | 1 << i], dp[msk] + cnt[i][-bal + SH]);
            }
          }
        }
      }
      smax(ans, dp[msk]);
    }
    cout << ans << endl;
  }
  