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 M = 1000000007;
  const int MM = 998244353;
  template <typename T, typename U>
  static inline void amin(T &x, U y) {
    if (y < x) x = y;
  }
  template <typename T, typename U>
  static inline void amax(T &x, U y) {
    if (x < y) x = y;
  }
  int _runtimeTerror_() {
    int n;
    cin >> n;
    vector<string> a(n);
    vector<int> len(n);
    for (int i = 0; i < n; ++i) {
      cin >> a[i];
      len[i] = (long long)a[i].size();
    }
    vector<map<int, int>> mp(n);
    vector<int> tsum(n, 0);
    for (int i = 0; i < n; ++i) {
      int mn = 0;
      for (char c : a[i]) {
        tsum[i] += c == '(';
        tsum[i] -= c == ')';
        if (tsum[i] <= mn) {
          ++mp[i][tsum[i]];
          amin(mn, tsum[i]);
        }
      }
    }
    vector<long long> dp(1 << n, -2000000000000000000);
    dp[0] = 0;
    vector<long long> sum(1 << n, 0);
    for (int i = 1; i < (1 << n); ++i) {
      for (int j = 0; j < n; ++j) {
        if (i & (1 << j)) {
          sum[i] += tsum[j];
        }
      }
    }
    long long ans = 0;
    for (int i = 0; i < (1 << n); ++i) {
      amax(ans, dp[i]);
      if (dp[i] == -2000000000000000000) {
        continue;
      }
      for (int j = 0; j < n; ++j) {
        if ((i & (1 << j)) == 0) {
          int f = mp[j].empty() ? 0 : mp[j].begin()->first;
          if (f + sum[i] < 0) {
            int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
            amax(ans, dp[i] + val);
          } else {
            int val = mp[j].count(-sum[i]) ? mp[j][-sum[i]] : 0;
            amax(dp[i | (1 << j)], dp[i] + val);
          }
        }
      }
    }
    cout << ans << "\n";
    return 0;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int TESTS = 1;
    while (TESTS--) _runtimeTerror_();
    return 0;
  }
  