浙工大姗姗杯round1 F

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1545    Accepted Submission(s): 686

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

34())(4()()6)))(((

 

Sample Output

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

题意:()(())这样左右配对的括号算好的，)(这样算不好的，如果交换一次能变成好的括号，就输出yes，不然就no，必须交换一次;

思路:参考MOOC数据结构堆栈对等式优先度的写法，模板题。坑点，因为必须交换一次，所以()是错的，输出no

#include <bits/stdc++.h>using namespace std;int main(){    stack<char> s;    int tt;    cin >> tt;    int n;    string p;    while(tt--)    {        cin >> n;        cin >> p;        for(int i = 0; i < n; i++)        {            if(p[i] == '(')                s.push(p[i]);            else {                if(s.size() > 0 && s.top() == '(')                    s.pop();                else {                    s.push(p[i]);                }            }        }        if( n%2 == 1||n < 2 || s.size() > 4)            cout << "No" << endl;        else if(n == 2 && s.size() == 0) cout << "No" << endl;        else cout << "Yes" << endl;        while(s.size())            s.pop();    }}