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 P = 1 << 20;
  int bad[21];
  int dp[P][2], le[P];
  int main() {
    int n;
    cin >> n;
    map<int, int> f[20];
    memset(dp, -0x3f, sizeof(dp));
    for (int i = 0; i < n; ++i) {
      string s;
      cin >> s;
      int sum = 0, mn = 0, cnt = 0;
      for (int j = 0; j < (int)s.size(); ++j) {
        if (s[j] == '(')
          ++sum;
        else
          --sum;
        mn = min(mn, sum);
        if (sum == mn) ++f[i][-sum];
        if (mn == 0 && sum == 0) ++cnt;
      }
      bad[i] = mn;
      le[1 << i] = sum;
      if (mn < 0)
        dp[1 << i][0] = cnt;
      else
        dp[1 << i][1] = cnt;
    }
    int p = (1 << n);
    for (int mask = 1; mask < p; ++mask) {
      for (int bit = 0; bit < n; ++bit) {
        if (mask & (1 << bit)) continue;
        dp[mask | (1 << bit)][0] = max(dp[mask | (1 << bit)][0], dp[mask][0]);
        int aux = 0;
        if (f[bit].count(le[mask])) aux = f[bit][le[mask]];
        if (le[mask] + bad[bit] >= 0)
          dp[mask | (1 << bit)][1] =
              max(dp[mask | (1 << bit)][1], dp[mask][1] + aux);
        else
          dp[mask | (1 << bit)][0] =
              max(dp[mask | (1 << bit)][0], dp[mask][1] + aux);
        le[mask | (1 << bit)] = le[mask] + le[1 << bit];
      }
    }
    printf("%d", max(dp[p - 1][0], dp[p - 1][1]));
    return 0;
  }
  