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;
  int NN = 20;
  string s[20];
  int cnt[20][(int)1e6];
  vector<int> tot(20, 0);
  vector<int> ts((1 << NN), 0);
  vector<int> vis((1 << NN), 0);
  vector<int> dp((1 << NN), 0);
  vector<int> lst((1 << NN), 0);
  int ans = 0;
  int fun(int x) {
    if (vis[x]) return dp[x];
    vis[x]++;
    if (ts[x] < 0) return 0;
    int cmax = 0;
    for (int i = 0; i < n; i++) {
      if (((1 << i) & x) == 0) {
        int y = cnt[i][(int)5e5 - ts[x]];
        if (lst[(1 << i)] + ts[x] >= 0) y += fun(x | (1 << i));
        cmax = max(cmax, y);
      }
    }
    dp[x] = cmax;
    return cmax;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> s[i];
      int curr = 0;
      bool less = false;
      int least = 0;
      for (auto j : s[i]) {
        if (j == '(')
          curr++;
        else
          curr--;
        if (curr <= 0 && least >= curr) cnt[i][(int)5e5 + curr]++;
        if (curr < 0) {
          least = min(least, curr);
        }
      }
      lst[(1 << i)] = least;
      tot[i] = curr;
    }
    int N = (1 << n);
    for (int i = 1; i < N; i++) {
      int a = 0;
      for (int j = 0; j < 20; j++) {
        if (((1 << j) & i) != 0) a += tot[j];
      }
      ts[i] = a;
    }
    fun(0);
    cout << dp[0];
  }
  