题目链接：

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

题意：

问交换两个位置能否使这些括号匹配;

思路：

看不能匹配的括号数,多于两对就NO了;有个wa点就是()直接wa哭我了;

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <bits/stdc++.h>
#include <stack>
#include <map>

using namespace std;

#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));

typedef  long long LL;

template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<‘0‘||CH>‘9‘;F= CH==‘-‘,CH=getchar());
    for(num=0;CH>=‘0‘&&CH<=‘9‘;num=num*10+CH-‘0‘,CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + ‘0‘);
    putchar(‘\n‘);
}

const LL mod=1e9+7;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=1e5+10;
const int maxn=5e3+4;
const double eps=1e-12;

int n;
char s[N];

int check()
{
    int num1=0,num2=0;
    For(i,1,n)
    {
        if(s[i]==‘(‘)num1++;
        else num2++;
    }
    if(num1!=num2)return 0;
    if(n==2)
    {
        if(s[1]==‘(‘&&s[2]==‘)‘)return 0;
        return 1;
    }
    num1=0;num2=0;
    For(i,1,n)
    {
        if(s[i]==‘(‘)num1++;
        else
        {
            if(num1==0)num2++;
            else num1--;
        }
    }
    if(num2>2)return 0;
    return 1;
}
int main()
{
   int T;
   read(T);
   while(T--)
   {
        read(n);
        scanf("%s",s+1);
        if(check())printf("Yes\n");
        else printf("No\n");
   }
    return 0;
}

AC代码：