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;
  string S[20];
  vector<pair<pair<int, int>, int> > psList[20];
  int ps[(1 << 20)], minps[(1 << 20)], dp[(1 << 20)];
  pair<int, int> Process(string &s, vector<pair<pair<int, int>, int> > &psList) {
    int ps = 0, minps = INT_MAX;
    for (int i = 0, n = s.size(); i < n; ++i) {
      ps += (s[i] == '(' ? 1 : -1);
      minps = min(minps, ps);
      psList.push_back({{ps, i}, i});
    }
    sort(psList.begin(), psList.end());
    int x = INT_MAX;
    for (auto &p : psList) {
      x = min(x, p.second);
      p.second = x;
    }
    return {ps, minps};
  }
  int Calculate(int i, int ps, int c) {
    ps = -ps;
    auto &v = psList[i];
    int k = lower_bound(v.begin(), v.end(), make_pair(make_pair(ps, 0), 0)) -
            v.begin();
    if (k >= v.size() || v[k].first.first != ps) {
      return c;
    }
    int n = k != 0 ? v[k - 1].second : INT_MAX;
    return c +
           (lower_bound(v.begin() + k, v.end(), make_pair(make_pair(ps, n), n)) -
            (v.begin() + k));
  }
  int main() {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; ++i) {
      cin >> S[i];
    }
    for (int i = 0; i < n; ++i) {
      int mask = (1 << i);
      tie(ps[mask], minps[mask]) = Process(S[i], psList[i]);
    }
    int m = (1 << n);
    ps[0] = 0, minps[0] = INT_MAX;
    for (int mask = 1; mask < m; ++mask) {
      int i = __builtin_ctz(mask);
      int imask = (1 << i);
      ps[mask] = ps[(mask ^ imask)] + ps[imask];
    }
    int ans = 0;
    for (int mask = 1; mask < m; ++mask) {
      int ansx = 0, minpsx = -1e6;
      for (int tmask = mask, i = 0; tmask != 0; tmask ^= (1 << i)) {
        i = __builtin_ctz(tmask);
        int rmask = mask ^ (1 << i);
        if (minps[rmask] < 0) {
          continue;
        }
        int ansy = Calculate(i, ps[rmask], dp[rmask]);
        ans = max(ans, ansy);
        int minpsy = min(minps[rmask], ps[rmask] + minps[(mask ^ rmask)]);
        minpsx = max(minpsx, minpsy);
        if (minpsy >= 0) {
          ansx = max(ansx, ansy);
        }
      }
      dp[mask] = ansx;
      minps[mask] = minpsx;
    }
    cout << ans << '\n';
    return 0;
  }
  