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;
  const int MAX = 4e5 + 5;
  int summ[20];
  int mmin[20];
  int maskSum[1 << 20];
  int cnt[MAX][20];
  int dp[1 << 20];
  int sol = 0;
  string arr[20];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n;
    cin >> n;
    for (int i = 0; i < n; i++) cin >> arr[i];
    for (int i = 0; i < (1 << n); i++) dp[i] = -1;
    for (int i = 0; i < n; i++) {
      string s = arr[i];
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '(')
          summ[i]++;
        else
          summ[i]--;
        mmin[i] = min(mmin[i], summ[i]);
        if (summ[i] == mmin[i]) cnt[-mmin[i]][i]++;
      }
    }
    dp[0] = 0;
    for (int mask = 0; mask < (1 << n); mask++) {
      if (dp[mask] < 0) continue;
      for (int j = 0; j < n; j++) {
        if (mask & (1 << j)) continue;
        int mask2 = mask | (1 << j);
        int val = dp[mask] + cnt[maskSum[mask]][j];
        sol = max(sol, val);
        maskSum[mask2] = maskSum[mask] + summ[j];
        if (maskSum[mask] + mmin[j] >= 0) {
          dp[mask2] = max(dp[mask2], val);
        }
      }
    }
    cout << sol << "\n";
    return 0;
  }
  