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 INF = 0x3f3f3f3f;
  unordered_map<int, int> cnt[20];
  unordered_map<int, int> pre[20];
  int sum[1 << 20];
  int mi[20];
  bool bad[1 << 20];
  int dp[1 << 20];
  inline int lowbit(int x) { return x & -x; }
  void solve() {
    int n;
    cin >> n;
    int ans = 0;
    for (int i = 0; i < n; i++) {
      string str;
      cin >> str;
      int t = 0;
      for (int j = 0, len = str.size(); j < len; j++) {
        t += (str[j] == '(' ? 1 : -1);
        mi[i] = min(mi[i], t);
        cnt[i][t]++;
        if (pre[i].find(t) == pre[i].end()) pre[i][t] = cnt[i][t + 1];
      }
      sum[1 << i] = t;
    }
    for (int i = 1; i < (1 << n); i++) {
      sum[i] = sum[i ^ lowbit(i)] + sum[lowbit(i)];
      bool ok = false;
      for (int j = 0; j < n; j++)
        if ((i >> j) & 1) {
          int k = (1 << j);
          if (dp[i ^ k] < 0) continue;
          int d = sum[i ^ k];
          if (d + mi[j] < 0) {
            ans = max(ans, dp[i ^ k] + pre[j][-d - 1]);
          } else {
            dp[i] = max(dp[i], dp[i ^ k] + cnt[j][-d]);
            ok = true;
          }
        }
      ans = max(ans, dp[i]);
      if (!ok) dp[i] = -1;
    }
    cout << ans << '\n';
  }
  signed main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    solve();
    return 0;
  }
  