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;
  int N;
  string S[20];
  int cnt[20], mc[20], ap[20];
  vector<pair<int, int> > vis[20];
  int sum[1 << 20];
  int dp[1 << 20];
  int main() {
    cin >> N;
    for (int i = 0; i < N; i++) {
      cin >> S[i];
      mc[i] = 0;
      for (char c : S[i]) {
        cnt[i] += c == '(' ? 1 : -1;
        if (mc[i] > cnt[i]) {
          vis[i].push_back(make_pair(mc[i], ap[i]));
          ap[i] = 0;
          mc[i] = cnt[i];
        }
        if (mc[i] == cnt[i]) ap[i]++;
      }
      vis[i].push_back(make_pair(mc[i], ap[i]));
      sort(vis[i].begin(), vis[i].end());
    }
    for (int i = 0; i < 1 << N; i++) dp[i] = -1;
    dp[0] = 0;
    int ans = 0;
    for (int i = 0; i < 1 << N; i++)
      if (dp[i] >= 0) {
        if (ans < dp[i]) ans = dp[i];
        int now = sum[i];
        for (int j = 0; j < N; j++)
          if (!(i >> j & 1)) {
            auto it =
                lower_bound(vis[j].begin(), vis[j].end(), make_pair(-now, 0));
            if (it != vis[j].end() && it->first == -now) {
              ans = max(ans, dp[i] + it->second);
            }
            if (now + mc[j] < 0) continue;
            int nxt = dp[i];
            if (now + mc[j] == 0) nxt += ap[j];
            int ni = i | 1 << j;
            sum[ni] = now + cnt[j];
            if (dp[ni] < nxt) dp[ni] = nxt;
          }
      }
    cout << ans << endl;
  }
  