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;
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<string> s(n);
    vector<int> bal(n);
    vector<int> mn(n);
    vector<unordered_map<int, int>> get(n);
    for (int i = 0; i < n; ++i) {
      cin >> s[i];
      int sum = 0;
      int mnn = 0;
      for (int j = 0; j < s[i].size(); ++j) {
        if (s[i][j] == '(') {
          ++sum;
        } else {
          --sum;
        }
        mnn = min(mnn, sum);
        if (mnn >= sum) {
          get[i][sum] += 1;
        }
      }
      bal[i] = sum;
      mn[i] = mnn;
    }
    vector<vector<int>> v(n, vector<int>(1 << n));
    vector<int> sum(1 << n);
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int j = 0; j < n; ++j) {
        if ((mask >> j) & 1) {
          sum[mask] = sum[mask ^ (1 << j)] + bal[j];
          break;
        }
      }
    }
    for (int i = 0; i < n; ++i) {
      for (int mask = 0; mask < (1 << n); ++mask) {
        int sz = sum[mask];
        if ((mask >> i) & 1) {
          sz -= bal[i];
          continue;
        }
        v[i][mask] = get[i][-sz];
      }
    }
    vector<vector<int>> dp(1 << n, vector<int>(2, -1e9));
    dp[0][0] = 0;
    for (int mask = 0; mask < (1 << n); ++mask) {
      for (int j = 0; j < n; ++j) {
        if ((mask >> j) & 1) {
          if (sum[mask ^ (1 << j)] + mn[j] >= 0) {
            dp[mask][0] =
                max(dp[mask][0], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
          } else {
            dp[mask][1] =
                max(dp[mask][1], dp[mask ^ (1 << j)][0] + v[j][mask ^ (1 << j)]);
          }
          dp[mask][1] = max(dp[mask][1], dp[mask ^ (1 << j)][1]);
        }
      }
    }
    cout << *max_element((dp[(1 << n) - 1]).begin(), (dp[(1 << n) - 1]).end());
  }
  