hdu5831

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1616    Accepted Submission(s): 710

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
思路：这题是指交换一次位置后仍然能够配对成功，其实出现一个右括号没有匹配是允许的，连续出现两个右括号没匹配也是可以的，但是三个及三个以上就不可以了，当两个括号时，（）时输出“No”的，)(输出“Yes”。
代码：
#include <iostream>#include <stdio.h>#include <string.h>#include <algorithm>#include <stack>using namespace std;char a[111111];int main(){    int T,n;    int s;    cin>>T;    while(T--)    {        cin>>n;        int flag=0;        int p=0;        int q=0;        for(int i=0;i<n;i++)            cin>>a[i];        for(int i=0;i<n;i++)        {            if(a[i]=='(')                p++;            if(a[i]==')')                q++;        }        if(p!=q)        {            cout<<"No"<<endl;            continue;        }        else        {            s=0;            if(p+q==2)            {                if(a[0]=='(')                    flag=1;            }            else            for(int i=0;i<n;i++)            {                if(a[i]=='(')                    s++;                if(a[i]==')')                    s--;                if(s==-3)                {                    flag=1;                    break;                }            }        }        if(flag==1)            cout<<"No"<<endl;        else            cout<<"Yes"<<endl;    }    return 0;}