【1】【栈模拟】HDU5831 Rikka with Parenthesis II

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)

Memory Limit: 65536/65536 K (Java/Others)

Problem Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Correct parentheses sequences can be defined recursively as follows:

1.The empty string “” is a correct sequence.

2.If “X” and “Y” are correct sequences, then “XY” (the concatenation of X and Y) is a correct sequence.

3.If “X” is a correct sequence, then “(X)” is a correct sequence.

Each correct parentheses sequence can be derived using the above rules.

Examples of correct parentheses sequences include “”, “()”, “()()()”, “(()())”, and “(((())))”.

Now Yuta has a parentheses sequence S, and he wants Rikka to choose two different position i,j and swap Si,Sj.

Rikka likes correct parentheses sequence. So she wants to know if she can change S to a correct parentheses sequence after this operation.

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number t(1<=t<=1000), the number of the testcases. And there are no more then 10 testcases with n>100

For each testcase, the first line contains an integers n(1<=n<=100000), the length of S. And the second line contains a string of length S which only contains ‘(’ and ‘)’.

Output

For each testcase, print “Yes” or “No” in a line.

Sample Input

3

4

())(

4

()()

6

)))(((

Sample Output

Yes

Yes

No

Hint

For the second sample input, Rikka can choose (1,3) or (2,4) to swap. But do nothing is not allowed.

链接

HDU5831

题意

给一个n长度的由“（”和“）”组成的字符串，然后交换其中两个位置上的括号，是否能够变成合法的括号串。

分析

16/8/11本来做的时候是直接模拟找括号对，然后排除特例排了好久。后来听学长讲他的方法，是一个Catalan模型，方法很妙，而且排除了之前那个做法的一个特例。即，求前缀和，前缀和<=-3则一定不成立。（自己领悟= =）Hint一定仔细看。