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 Nmax = 20;
  int n;
  string s[20];
  int sum[20], minl[20];
  int f[1 << 20];
  int g[20][800010];
  int main() {
    ios ::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n;
    for (int i = 0; i < n; ++i) cin >> s[i];
    for (int i = 0; i < n; ++i) {
      minl[i] = 1e9;
      for (auto c : s[i]) {
        sum[i] += (c == '(' ? 1 : -1);
        minl[i] = min(minl[i], sum[i]);
        if (sum[i] > minl[i]) continue;
        ++g[i][sum[i] + s[i].size()];
      }
    }
    memset(f, -1, sizeof(f));
    f[0] = 0;
    int ans = 0;
    for (int mask = 0; mask < (1 << n); ++mask) {
      if (f[mask] < 0) continue;
      int now = 0;
      for (int i = 0; i < n; ++i)
        if (mask >> i & 1) now += sum[i];
      for (int i = 0; i < n; ++i) {
        if (mask >> i & 1) continue;
        int bonus = (now > s[i].size() ? 0 : g[i][-now + s[i].size()]);
        ans = max(ans, f[mask] + bonus);
        if (now + minl[i] < 0) continue;
        f[mask | (1 << i)] = max(f[mask | (1 << i)], f[mask] + bonus);
      }
    }
    cout << ans << '\n';
  }
  