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;
  inline int read() {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  inline long long llread() {
    long long x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= ch == '-', ch = getchar();
    while (isdigit(ch)) x = 10 * x + ch - '0', ch = getchar();
    return f ? -x : x;
  }
  int n;
  char s[25][400010];
  int l[25], rest[25];
  vector<int> v[25][400010];
  int dp[(1 << 20) + 10][2], g[(1 << 20) + 10];
  signed main() {
    n = read();
    for (int i = (1); i <= (n); i++) scanf("%s", s[i] + 1);
    for (int i = (1); i <= (n); i++) l[i] = strlen(s[i] + 1);
    for (int i = (1); i <= (n); i++) {
      int top = 0;
      for (int j = (1); j <= (l[i]); j++) {
        if (s[i][j] == '(')
          top++;
        else
          top--;
        if (top <= 0) v[i][-top].push_back(j);
      }
      rest[i] = top;
    }
    for (int i = (1); i <= ((1 << n) - 1); i++) dp[i][0] = dp[i][1] = -0x3f3f3f3f;
    dp[0][0] = dp[0][1] = g[0] = 0;
    for (int mask = (1); mask <= ((1 << n) - 1); mask++) {
      for (int i = (1); i <= (n); i++) {
        if (mask & (1 << (i - 1))) {
          int S = mask ^ (1 << (i - 1));
          if (dp[mask][0] < dp[S][0]) {
            dp[mask][0] = dp[S][0];
          };
          if (g[S] >= 0) {
            int x;
            if (v[i][g[S] + 1].size() == 0)
              x = l[i] + 1;
            else
              x = v[i][g[S] + 1][0];
            int y = lower_bound(v[i][g[S]].begin(), v[i][g[S]].end(), x) -
                    v[i][g[S]].begin();
            if (x == l[i] + 1) {
              if (dp[mask][1] < dp[S][1] + y) {
                dp[mask][1] = dp[S][1] + y;
              };
            }
            if (dp[mask][0] < dp[S][1] + y) {
              dp[mask][0] = dp[S][1] + y;
            };
          }
          g[mask] = g[S] + rest[i];
        }
      }
    }
    printf("%d\n", max(dp[(1 << n) - 1][0], dp[(1 << n) - 1][1]));
    return 0;
  }
  