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, N[22], cnt[22];
  char s[400004];
  int dp[1048600];
  int mn[22];
  vector<int> f[22];
  vector<int> z[800006];
  int main() {
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      int m;
      scanf("%s", s + 1), m = strlen(s + 1);
      int sum = 0;
      for (int j = 1; j <= m; j++) {
        sum += (s[j] == '(' ? 1 : -1);
        z[sum + m].push_back(j);
        mn[i] = min(mn[i], sum);
      }
      f[i].resize(m + 10);
      for (int j = 0; j <= m; j++) {
        f[i][j] = 0;
        int nxt = 1e9;
        if (j > 0 && !z[j - 1].empty()) nxt = z[j - 1][0];
        if (!z[j].empty())
          for (auto x : z[j]) {
            if (x > nxt) break;
            f[i][j]++;
          }
      }
      for (int j = 0; j <= m + m; j++) z[j].clear();
      N[i] = m, cnt[i] = sum;
    }
    int ans = 0;
    for (int i = 1; i < (1 << n); i++) dp[i] = -1e9;
    for (int i = 0; i < (1 << n); i++) {
      int cur = 0;
      for (int j = 0; j < n; j++)
        if ((1 << j) & i) cur += cnt[j];
      for (int j = 0; j < n; j++)
        if (!((1 << j) & i)) {
          int MN = mn[j];
          int tmp = dp[i];
          if (N[j] >= cur && cur >= 0) tmp += f[j][N[j] - cur];
          ans = max(tmp, ans);
          if (cur + MN >= 0) dp[i | (1 << j)] = max(dp[i | (1 << j)], tmp);
        }
    }
    printf("%d", ans);
  }
  