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 inf = 1e9, N = 25, M = 4e5 + 5, S = 4e6 + 5;
  struct point {
    int w, sum;
    vector<int> t;
  } a[N];
  inline int read() {
    int F = 1, ANS = 0;
    char C = getchar();
    while (C < '0' || C > '9') {
      if (C == '-') F = -1;
      C = getchar();
    }
    while (C >= '0' && C <= '9') {
      ANS = ANS * 10 + C - '0';
      C = getchar();
    }
    return F * ANS;
  }
  char s[M];
  int dp[S], ans;
  inline char readchar() {
    char C = getchar();
    while (C < 33 || C > 126) C = getchar();
    return C;
  }
  inline int raed() { return read(); }
  int main() {
    n = read();
    for (int i = 0; i < n; i++) {
      scanf("%s", s + 1);
      int len = strlen(s + 1);
      int tot = 0, res = 0;
      for (int j = 0; j <= len; j++) a[i].t.push_back(0);
      for (int j = 1; j <= len; j++) {
        if (s[j] == '(')
          tot++;
        else
          tot--;
        if (tot <= res) {
          res = tot;
          a[i].t[-tot]++;
        }
      }
      a[i].w = -res;
      a[i].sum = tot;
    }
    for (int i = 1; i <= (1 << n) - 1; i++) dp[i] = -inf;
    dp[S] = 0;
    for (int i = 1; i <= (1 << n) - 1; i++) {
      int sum = 0;
      for (int j = 0; j < n; j++) {
        if (((1 << j) & i) == 0) continue;
        sum += a[j].sum;
      }
      for (int j = 0; j < n; j++) {
        if (((1 << j) & i) == 0) continue;
        sum -= a[j].sum;
        int k = ((1 << j) ^ i);
        if (sum < 0) {
          sum += a[j].sum;
          continue;
        }
        if (dp[k] < 0) {
          sum += a[j].sum;
          continue;
        }
        if (sum > a[j].w) {
          dp[i] = max(dp[i], dp[k]);
          sum += a[j].sum;
          continue;
        }
        if (sum == a[j].w) {
          dp[i] = max(dp[i], dp[k] + a[j].t[sum]);
          sum += a[j].sum;
          continue;
        }
        ans = max(ans, dp[k] + a[j].t[sum]);
        sum += a[j].sum;
      }
      ans = max(ans, dp[i]);
    }
    cout << ans << endl;
    return 0;
  }
  