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A14595. RBS

编程题 普及/提高-

题目描述

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 $s_1, s_2, \dots, s_n$ . 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 $s_1 + s_2 + \dots + s_n$ has as many non-empty prefixes that are RBS as possible.

输入格式

The first line contains a single integer $n$ ( $1 \le n \le 20$ ).

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 $4 \cdot 10^5$ .

输出格式

Print one integer — the maximum number of non-empty prefixes that are RBS for the string $s_1 + s_2 + \dots + s_n$ , if the strings $s_1, s_2, \dots, s_n$ can be rearranged arbitrarily.

输入输出样例

输入 #1
2
(
)
输出 #1
1
输入 #2
4
()()())
(
(
)
输出 #2
4
输入 #3
1
(())
输出 #3
1
输入 #4
1
)(()
输出 #4
0

说明/提示

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.
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