A15181 | Recover an RBS
时间限制1s
内存限制256MB
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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.
There was an RBS. Some brackets have been replaced with question marks. Is it true that there is a unique way to replace question marks with brackets, so that the resulting sequence is an RBS?
- bracket sequences "()()" and "(())" are regular (the resulting expressions are: "(1)+(1)" and "((1+1)+1)");
- bracket sequences ")(", "(" and ")" are not.
There was an RBS. Some brackets have been replaced with question marks. Is it true that there is a unique way to replace question marks with brackets, so that the resulting sequence is an RBS?
输入格式
The first line contains a single integer $t$ ( $1 \le t \le 5 \cdot 10^4$ ) — the number of testcases.
The only line of each testcase contains an RBS with some brackets replaced with question marks. Each character is either '(', ')' or '?'. At least one RBS can be recovered from the given sequence.
The total length of the sequences over all testcases doesn't exceed $2 \cdot 10^5$ .
The only line of each testcase contains an RBS with some brackets replaced with question marks. Each character is either '(', ')' or '?'. At least one RBS can be recovered from the given sequence.
The total length of the sequences over all testcases doesn't exceed $2 \cdot 10^5$ .
输出格式
For each testcase, print "YES" if the way to replace question marks with brackets, so that the resulting sequence is an RBS, is unique. If there is more than one way, then print "NO".
输入输出样例
输入 #1
5 (?)) ?????? () ?? ?(?)()?)
输出 #1
YES NO YES YES NO
In the first testcase, the only possible original RBS is "(())".
In the second testcase, there are multiple ways to recover an RBS.
In the third and the fourth testcases, the only possible original RBS is "()".
In the fifth testcase, the original RBS can be either "((()()))" or "(())()()".
In the second testcase, there are multiple ways to recover an RBS.
In the third and the fourth testcases, the only possible original RBS is "()".
In the fifth testcase, the original RBS can be either "((()()))" or "(())()()".
C++ 编辑器
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评测结果:Accepted