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 long long MOD = 1e9 + 7;
  const int bas = 5e5;
  int n, cnt[30][1000000], mnv[30], add[30], sum[1100000], dp[1100000], ans = 0;
  string s[30];
  int main() {
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      int fim, mn = 1e9, cur = 0;
      for (int j = 0; j < s[i].size(); ++j) {
        cur += (s[i][j] == '(' ? 1 : -1);
        if (cur < mn) mn = cur, fim = j;
      }
      add[i] = cur;
      mnv[i] = mn;
      cur = 0;
      mn = 1e9;
      for (int j = 0; j < s[i].size(); ++j) {
        cur += (s[i][j] == '(' ? 1 : -1);
        if (mn >= cur) ++cnt[i][bas + cur];
        mn = min(mn, cur);
      }
    }
    for (int i = 1; i <= (1 << n) - 1; ++i) {
      int nn = __builtin_ctz(i);
      sum[i] = add[nn] + sum[i ^ (1 << nn)];
    }
    for (int i = 0; i < (1 << n) + 5; ++i) dp[i] = -1;
    dp[0] = 0;
    for (int i = 0; i < 1 << n; ++i)
      if (dp[i] > -1) {
        ans = max(ans, dp[i]);
        int cur = sum[i];
        for (int j = 0; j < n; ++j)
          if ((i & (1 << j)) == 0) {
            if (cur + mnv[j] < 0) {
              ans = max(ans, dp[i] + cnt[j][-cur + bas]);
              continue;
            }
            dp[i | (1 << j)] = max(dp[i | (1 << j)], dp[i] + cnt[j][-cur + bas]);
          }
      }
    cout << ans << endl;
    return 0;
  }
  