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;
  const int MAXN = 4e5 + 7;
  char s[MAXN];
  int g[20][MAXN], w[20], mn[20], f[1 << 20], W[1 << 20], ans;
  int main() {
    memset(f, -0x3f, sizeof(f));
    f[0] = 0;
    scanf("%d", &n);
    for (int t = 0; t < n; ++t) {
      scanf("%s", s + 1);
      int cnt = 0, m = strlen(s + 1);
      for (int j = 1; j <= m; ++j) {
        if (s[j] == '(')
          ++cnt;
        else
          --cnt;
        if (cnt <= mn[t]) g[t][-cnt]++, mn[t] = cnt;
      }
      w[t] = cnt;
    }
    for (int s = 0; s < (1 << n); ++s) {
      if (f[s] < 0) continue;
      for (int i = 0; i < n; ++i) {
        if (s >> i & 1) continue;
        int t = s ^ (1 << i);
        ans = max(ans, f[s] + g[i][W[s]]);
        if (W[s] + mn[i] >= 0) {
          W[t] = W[s] + w[i];
          f[t] = max(f[t], f[s] + g[i][W[s]]);
        }
      }
    }
    printf("%d\n", ans);
  }
  