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, sum[20], dp[1 << 20][2], a[1 << 20];
  map<int, vector<int>> pref[20];
  map<int, int> stop[20];
  int main() {
    cin >> N;
    for (int i = 0; i < N; ++i) {
      string s;
      cin >> s;
      for (int j = 0; j < s.size(); ++j) {
        pref[i][sum[i] += s[j] == '(' ? 1 : -1].push_back(j);
        if (!stop[i].count(sum[i])) stop[i][sum[i]] = j;
      }
      for (int j = stop[i].begin()->first + 1; j <= prev(stop[i].end())->first;
           ++j) {
        stop[i][j] = min(stop[i].count(j) ? stop[i][j] : N, stop[i][j - 1]);
      }
    }
    for (int i = 0; i < (1 << N); ++i) dp[i][1] = -1e9;
    dp[0][1] = 0;
    int ans = 0;
    for (int i = 1; i < (1 << N); ++i) {
      for (int j = 0; j < N; ++j) {
        if (~i & (1 << j)) continue;
        int b = pref[j].begin()->first + a[i ^ (1 << j)] >= 0;
        a[i] = sum[j] + a[i ^ (1 << j)];
        int s = a[i ^ (1 << j)];
        dp[i][b] = max(
            dp[i][b],
            dp[i ^ (1 << j)][1] +
                (b ? (int)(pref[j].count(-s) ? pref[j][-s].size() : 0)
                   : (int)(pref[j].count(-s)
                               ? lower_bound(pref[j][-s].begin(),
                                             pref[j][-s].end(), stop[j][-s - 1]) -
                                     pref[j][-s].begin()
                               : 0)));
      }
      ans = max(ans, dp[i][0] = max(dp[i][0], dp[i][1]));
    }
    cout << ans << endl;
    return 0;
  }
  