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 maxn = 21, maxv = 400005;
  int n, m, ans, tot;
  int f[1 << maxn], mn[maxn], cnt[maxn], sum[1 << maxn], w[maxn][maxv << 1],
      ok[maxn][maxv << 1];
  string s;
  int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      int now = 0;
      cin >> s;
      for (int j = 0; j < s.size(); j++) {
        if (s[j] == '(') now++;
        if (s[j] == ')') ok[i][400000 + now] = 1, now--;
        if (ok[i][400000 + now] == 0) w[i][400000 + now]++;
        mn[i] = min(mn[i], now);
      }
      sum[1 << (i - 1)] = now;
    }
    memset(f, -1, sizeof(f)), f[0] = 0;
    for (int i = 1; i < (1 << n); i++) {
      sum[i] = sum[i ^ i & -i] + sum[i & -i];
      for (int j = 1; j <= n; j++)
        if ((i >> (j - 1)) & 1) {
          int pre = (i ^ (1 << (j - 1)));
          if (f[pre] == -1) continue;
          if (sum[pre] + mn[j] >= 0)
            f[i] = max(f[i], f[pre] + w[j][400000 - sum[pre]]);
          else
            ans = max(ans, f[pre] + w[j][400000 - sum[pre]]);
        }
      ans = max(ans, f[i]);
    }
    printf("%d\n", ans);
    return 0;
  }
  