CF3D.Least Cost Bracket Sequence
提高+/省选-
通过率:0%
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题目描述
This is yet another problem on regular bracket sequences.
A bracket sequence is called regular, if by inserting "+" and "1" into it we get a correct mathematical expression. For example, sequences "(())()", "()" and "(()(()))" are regular, while ")(", "(()" and "(()))(" are not. You have a pattern of a bracket sequence that consists of characters "(", ")" and "?". You have to replace each character "?" with a bracket so, that you get a regular bracket sequence.
For each character "?" the cost of its replacement with "(" and ")" is given. Among all the possible variants your should choose the cheapest.
给一个序列,序列里面会有左括号、问号、右括号。对于一个‘?’而言,可以将其替换为一个‘(’,也可以替换成一个‘)’,但是都有相应的代价。问:如何替换使得代价最小。前提是替换之后的序列中,括号是匹配的。如果不能替换为一个括号匹配的序列则输出-1。
输入格式
The first line contains a non-empty pattern of even length, consisting of characters "(", ")" and "?". Its length doesn't exceed 5⋅104 . Then there follow m lines, where m is the number of characters "?" in the pattern. Each line contains two integer numbers ai and bi ( 1<=ai,bi<=106 ), where ai is the cost of replacing the i -th character "?" with an opening bracket, and bi — with a closing one.
第一行是序列,序列长度不超过 5×104,下面m(m是?的数量)行有每行 2 个数据,第一个是 '('的代价,第2个是 ')' 的代价。
输出格式
Print the cost of the optimal regular bracket sequence in the first line, and the required sequence in the second.
Print -1, if there is no answer. If the answer is not unique, print any of them.
第一行输出最小代价,第二行输出替换后的序列。不行输出 −1。
输入输出样例
输入#1
(??) 1 2 2 8
输出#1
4 ()()