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;
  string s[21];
  int mp[21][400005];
  int baki[21];
  int mx[21];
  int dp[1 << 21];
  int run(int mask) {
    if (mask == (1 << n) - 1) {
      return 0;
    }
    if (dp[mask] != -1) return dp[mask];
    int a = 0;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) {
        a += baki[i] - mx[i];
      }
    }
    int tot = 0;
    for (int i = 0; i < n; i++) {
      if (mask & (1 << i)) continue;
      int ans = 0;
      ans += mp[i][a];
      if (mx[i] <= a) {
        ans += run(mask | (1 << i));
      }
      tot = max(ans, tot);
    }
    return dp[mask] = tot;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ;
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> s[i];
      int a = 0, b = 0;
      int ex = 0;
      for (auto u : s[i]) {
        if (u == '(') {
          a++;
        } else {
          if (a == 0) {
            ex++;
            mp[i][ex]++;
          } else {
            a--;
            if (a == 0) mp[i][ex]++;
          }
        }
        mx[i] = max(mx[i], ex);
      }
      baki[i] = a;
    }
    memset(dp, -1, sizeof dp);
    cout << run(0);
  }
  