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>
  const int N = 400005;
  const int M = 21;
  std::vector<int> val[M], cnt[M], mn[M], pos[M];
  int cnt1[1 << (M - 1)], f[1 << (M - 1)][2];
  bool valid[1 << (M - 1)];
  inline int lb(int x) { return x & (-x); }
  void init(int x) {
    static char s[N];
    static int pre[N << 1];
    scanf("%s", s + 1);
    int sz = strlen(s + 1), sum = 0;
    for (int i = 1; i <= sz; i++) {
      if (s[i] == '(')
        sum++;
      else
        sum--;
      val[x].push_back(sum);
      if (!mn[x].size())
        mn[x].push_back(sum);
      else
        mn[x].push_back(std::min(mn[x].back(), sum));
      if (mn[x].back() == val[x].back())
        pos[x].push_back(i - 1);
      else
        pos[x].push_back(pos[x].back());
    }
    for (int i = 0; i < val[x].size(); i++) {
      cnt[x].push_back(++pre[val[x][i] + N - 5]);
    }
    for (int i = 0; i < val[x].size(); i++) pre[val[x][i] + N - 5] = 0;
  }
  int calc(int x, int k) {
    int l = 0, r = mn[x].size() - 1, ans = 0;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (mn[x][mid] >= k)
        ans = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    return mn[x][ans] == k ? cnt[x][pos[x][ans]] : 0;
  }
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) init(i);
    for (int i = 1; i <= n; i++) cnt1[1 << (i - 1)] = val[i].back();
    for (int i = 1; i < (1 << n); i++) cnt1[i] = cnt1[i ^ lb(i)] + cnt1[lb(i)];
    for (int i = 0; i < (1 << n); i++) {
      for (int j = 0; j < n; j++) {
        if (!((i >> j) & 1)) {
          f[i ^ (1 << j)][1] = std::max(f[i][1], f[i ^ (1 << j)][1]);
          if (valid[i] || i == 0) {
            if (cnt1[i] + mn[j + 1].back() >= 0)
              f[i ^ (1 << j)][0] =
                  std::max(f[i ^ (1 << j)][0], f[i][0] + calc(j + 1, -cnt1[i])),
                          valid[i ^ (1 << j)] = 1;
            else
              f[i ^ (1 << j)][1] =
                  std::max(f[i ^ (1 << j)][1], f[i][0] + calc(j + 1, -cnt1[i]));
          }
        }
      }
    }
    printf("%d\n", std::max(f[(1 << n) - 1][0], f[(1 << n) - 1][1]));
  }
  