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 MAXN = (1ll << 21);
  long long sum[MAXN], mins[22];
  vector<vector<long long>> vm;
  long long dp[MAXN];
  long long f(long long x) {
    long long &res = dp[x];
    if (res >= 0) return res;
    if (x == (1ll << ((int)vm.size())) - 1) return res = 0;
    res = 0;
    long long s = sum[x];
    for (int i = 0, ggdem = ((int)vm.size()); i < ggdem; ++i) {
      if (!((1ll << i) & x)) {
        long long val = 0;
        if (mins[i] + s <= 0) {
          val += vm[i][s];
        }
        if (mins[i] + s >= 0) {
          val += f(x ^ (1ll << i));
        }
        res = max(res, val);
      }
    }
    return res;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long n;
    cin >> n;
    vector<string> s(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) cin >> s[i];
    vm.resize(n);
    for (int i = 0, ggdem = n; i < ggdem; ++i) {
      long long mini = 0, va = 0;
      vm[i] = {0};
      for (auto j : s[i]) {
        if (j == '(')
          va++;
        else
          va--;
        if (va == mini) vm[i].back()++;
        if (va < mini) vm[i].push_back(1);
        mini = min(mini, va);
      }
      mins[i] = mini;
      sum[(1ll << i)] = va;
    }
    for (int i = 1, ggdem = (1ll << n); i < ggdem; ++i) {
      for (int j = 0, ggdem = n; j < ggdem; ++j) {
        if ((1ll << j) & i) {
          sum[i] = sum[i ^ (1ll << j)] + sum[(1ll << j)];
        }
      }
    }
    memset((dp), (-1), sizeof(dp));
    cout << f(0) << "\n";
    return 0;
  }
  