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>
  int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(nullptr);
    int n;
    std::cin >> n;
    std::vector<int> m(n);
    std::vector<std::vector<int>> mn(n), c(n);
    std::vector<int> d(n);
    for (int i = 0; i < n; ++i) {
      std::string s;
      std::cin >> s;
      m[i] = s.size();
      int now = 0;
      mn[i].resize(m[i] + 1), c[i].resize(m[i] + 1);
      for (int j = 0; j < m[i]; ++j) {
        if (s[j] == '(') {
          ++now;
        } else {
          --now;
        }
        if (now < mn[i][j]) {
          mn[i][j + 1] = now;
          c[i][j + 1] = 1;
        } else if (mn[i][j] == now) {
          mn[i][j + 1] = now;
          c[i][j + 1] = c[i][j] + 1;
        } else {
          mn[i][j + 1] = mn[i][j];
          c[i][j + 1] = c[i][j];
        }
      }
      d[i] = now;
    }
    std::vector<int> sum(1 << n);
    for (int i = 0; i < n; ++i) {
      sum[1 << i] = d[i];
    }
    for (int S = 1; S < (1 << n); ++S) {
      sum[S] = sum[S & (S - 1)] + sum[S & -S];
    }
    std::vector<int> f(1 << n, -1);
    f[0] = 0;
    for (int S = 1; S < (1 << n); ++S) {
      for (int i = 0; i < n; ++i) {
        int v = sum[S ^ (1 << i)] + mn[i][m[i]];
        if (S >> i & 1 && f[S ^ (1 << i)] >= 0 && v >= 0) {
          f[S] = std::max(f[S], f[S ^ (1 << i)] + (v == 0 ? c[i][m[i]] : 0));
        }
      }
    }
    int ans = 0;
    for (int S = 0; S < (1 << n); ++S) {
      if (f[S] >= 0) {
        ans = std::max(ans, f[S]);
        for (int i = 0; i < n; ++i) {
          if (!(S >> i & 1)) {
            int p = std::upper_bound(mn[i].begin(), mn[i].end(), -sum[S],
                                     std::greater<int>()) -
                    mn[i].begin() - 1;
            ans = std::max(ans, f[S] + (mn[i][p] == -sum[S] ? c[i][p] : 0));
          }
        }
      }
    }
    std::cout << ans << "\n";
  }
  