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, a[22], dp[1 << 21], sum[1 << 21], ans = 0;
  char s[500005];
  vector<int> pos[21];
  vector<vector<int>> c[21];
  void solve(int x, int id) {
    if (dp[x] == -1) return;
    int p = 0;
    if (pos[id].size() > sum[x] + 1) p = pos[id][sum[x] + 1];
    if (!p) {
      int k = 0;
      if (c[id].size() > sum[x]) k = c[id][sum[x]].size();
      dp[x | (1 << id - 1)] = max(dp[x | (1 << id - 1)], dp[x] + k);
      ans = max(ans, dp[x | (1 << id - 1)]);
    } else {
      if (c[id].size() <= sum[x]) return;
      int k = upper_bound(c[id][sum[x]].begin(), c[id][sum[x]].end(), p) -
              c[id][sum[x]].begin();
      ans = max(ans, dp[x] + k);
    }
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    memset(dp, -1, sizeof(dp));
    dp[0] = 0;
    cin >> n;
    for (int i = 1, k = 0; i <= n; ++i, k = 0) {
      cin >> s + 1;
      int len = strlen(s + 1);
      pos[i].resize(len + 1);
      c[i].resize(len + 1);
      for (int j = 1; j <= len; ++j) {
        k += (s[j] == '(' ? 1 : -1);
        if (k <= 0) {
          if (!pos[i][-k]) pos[i][-k] = j;
          c[i][-k].push_back(j);
        }
      }
      a[i] = k;
    }
    for (int i = 1; i < (1 << n); ++i) {
      int j = __builtin_ffs(i);
      sum[i] = sum[i ^ (1 << j - 1)] + a[j];
    }
    for (int i = 1; i < (1 << n); ++i)
      for (int j = 1; j <= i; j <<= 1)
        if (i & j) solve(i ^ j, __builtin_ffs(j));
    cout << ans << '\n';
    return 0;
  }
  