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;
  char S[23][400005];
  vector<int> zhi[23];
  int vvv[23];
  int dp[(1 << 20) + 3];
  inline int countOne(const vector<int>& zhi, int x) {
    auto iter1 = lower_bound(zhi.begin(), zhi.end(), x);
    auto iter2 = upper_bound(zhi.begin(), zhi.end(), x);
    int ret = iter2 - iter1;
    if (x == 0) --ret;
    return ret;
  }
  int main() {
    scanf("%d", &N);
    for (int i = 0; i < N; ++i) {
      scanf("%s", S[i]);
      int val = 0;
      zhi[i].push_back(-val);
      for (int j = 0; S[i][j]; ++j) {
        if (S[i][j] == '(')
          ++val;
        else
          --val;
        if (val <= -*zhi[i].rbegin()) zhi[i].push_back(-val);
      }
      vvv[i] = val;
    }
    int ans = 0;
    memset(dp, -1, sizeof(dp));
    dp[0] = 0;
    for (int f = 1; f < (1 << N); ++f) {
      int s = 0;
      for (int i = 0; i < N; ++i)
        if (f & (1 << i)) s += vvv[i];
      for (int i = 0; i < N; ++i)
        if ((f & (1 << i)) && dp[f ^ (1 << i)] != -1) {
          int tmp = dp[f ^ (1 << i)];
          int now = tmp + countOne(zhi[i], s - vvv[i]);
          if (-*zhi[i].rbegin() + s - vvv[i] < 0) {
            ans = max(ans, now);
            continue;
          }
          dp[f] = max(dp[f], now);
        }
    }
    ans = max(ans, dp[(1 << N) - 1]);
    printf("%d\n", ans);
    return 0;
  }
  