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 = 1e6;
  const int oo = 1e9;
  const int smax = 4e5;
  int n, m, num, d[21], c[21], pow2[21], f[1048580][2], sum[1048580];
  bool kt;
  string s;
  pair<int, int> goes[21][400001], temp, tam;
  vector<int> v[smax];
  pair<int, int> go(int i, int flag, int vaule) {
    if (flag) return {0, 1};
    return goes[i][vaule];
  }
  void maximize(int *x, int y) {
    if (*x < y) *x = y;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    memset(f, -1, sizeof(f));
    pow2[0] = 1;
    for (int i = 1; i <= 20; i++) pow2[i] = pow2[i - 1] * 2;
    cin >> n;
    for (int j = 1; j < pow2[n]; j++) f[j][0] = f[j][1] = -oo;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j <= smax; j++) v[j].clear();
      cin >> s;
      m = s.size();
      c[i] = d[i] = 0;
      kt = true;
      for (int j = 0; j < m; j++) {
        if (s[j] == '(')
          d[i]++;
        else
          d[i]--;
        if (d[i] < c[i]) c[i] = d[i];
        if (d[i] <= 0) v[-d[i]].push_back(j);
      }
      for (int j = 0; j <= smax; j++) {
        if (v[j].empty())
          num = 0;
        else {
          if (!v[j + 1].empty())
            num =
                upper_bound(v[j].begin(), v[j].end(), v[j + 1][0]) - v[j].begin();
          else
            num = v[j].size();
        }
        if (c[i] + j < 0)
          goes[i][j] = {num, 1};
        else
          goes[i][j] = {num, 0};
      }
      for (int j = 1; j < pow2[n]; j++)
        if ((j >> (i - 1)) & 1) sum[j] += d[i];
    }
    f[0][0] = sum[0] = 0;
    f[0][1] = -oo;
    for (int i = 0; i < pow2[n]; i++) {
      for (int j = 1; j <= n; j++)
        if (!((i >> (j - 1)) & 1)) {
          if (f[i][1] != -oo) maximize(&f[i ^ pow2[j - 1]][1], f[i][1]);
          if (f[i][0] == -oo) continue;
          temp = go(j, 0, sum[i]);
          maximize(&f[i ^ pow2[j - 1]][temp.second], f[i][0] + temp.first);
        }
    }
    cout << max(f[pow2[n] - 1][0], f[pow2[n] - 1][1]);
    return 0;
  }
  