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 f[1 << 20][2], sum[1 << 20], val[20], a[20], b[20];
  string s[20];
  vector<int> vec[20], pre[20];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    int n;
    cin >> n;
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      int m = s[i].size(), cnt = 0;
      pre[i].resize(m, 0);
      for (int j = 0; j < m; ++j) {
        if (s[i][j] == '(')
          ++cnt;
        else {
          ++b[i];
          if (cnt == 0)
            vec[i].push_back(j);
          else
            --cnt;
        }
        if (cnt == 0) pre[i][j]++;
        if (j) pre[i][j] += pre[i][j - 1];
      }
      for (int j = 0; j < m; ++j)
        if (s[i][j] == '(')
          ++a[i];
        else
          --a[i];
    }
    int all = (1 << n) - 1;
    for (int i = 1; i <= all; ++i)
      for (int j = 0; j < n; ++j)
        if (i >> j & 1) sum[i] += a[j];
    f[0][1] = -1e9;
    for (int S = 1; S <= all; ++S) {
      f[S][0] = f[S][1] = -1e9;
      for (int i = 0; i < n; ++i)
        if (S >> i & 1) {
          int T = S ^ (1 << i);
          if (sum[T] >= vec[i].size()) {
            f[S][0] = max(f[S][0], f[T][0]);
            if (sum[T] == vec[i].size()) {
              if (vec[i].size() == 0)
                f[S][0] = max(f[S][0], f[T][0] + pre[i].back());
              else {
                int l = vec[i].back(), r = s[i].size() - 1;
                f[S][0] = max(f[S][0], f[T][0] + 1 + pre[i][r] - pre[i][l]);
              }
            }
          }
          f[S][1] = max(f[S][1], f[T][1]);
          if (sum[T] >= 0 && sum[T] < vec[i].size()) {
            f[S][1] = max(f[S][1], f[T][0]);
            int r = vec[i][sum[T]] - 1;
            if (r < 0) continue;
            if (sum[T] == 0) {
              f[S][1] = max(f[S][1], f[T][0] + pre[i][r]);
            } else {
              int l = vec[i][sum[T] - 1];
              f[S][1] = max(f[S][1], f[T][0] + pre[i][r] - pre[i][l] + 1);
            }
          }
        }
    }
    cout << max(f[all][0], f[all][1]);
  }
  