hdu 5831 Rikka with Parenthesis II【水题】

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0


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.
 

题目大意：

给你n长度的一个括号串，问你是否能够在必须交换两个括号的情况下，使得最终交换得到的括号串是匹配的。

思路：

1、首先，需要明确一点，如果n%2==1，输出No，括号中的左括号和右括号的数目不同，输出No。因为必须挪动，所以我们n==2的时候要特判一下。

2、如果n%2==0&&左括号数目==右括号数目，那么我们遍历一遍字符串，如果遇到左括号，sum++，记录左括号的数目.如果遇到右括号，如果sum大于0，那么sum--。否则讨论：从第一个位子到当前位子作为一个子串的话，其子串一定做不到所有括号都匹配，所以这个括号是一定需要挪动的，那么我们问题转化：统计这样必须挪动的括号的个数。

3、那么对于这种需要挪动的括号进行记数，如果这样的括号有0或者1或者2个，都是可行的:

①0个需要挪动的括号，而且n>=4，那么随便挪动一对括号就行。

②1个需要挪动的括号，那么一定有一个右括号也需要挪动与之匹配：）（

③2个需要挪动的括号，那么一定有两个右括号也需要挪动与之匹配：））（（挪动14就可以达到目的.

Ac代码：

#include<stdio.h>
#include<string.h>
#include<stack>
#include<iostream>
using namespace std;
char a[1000000];
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
stack<int >s;
int n;
scanf("%d",&n);
cin>>a;
if(n==2)
{
if(a[0]==')'&&a[1]=='(')
{
printf("Yes\n");
}
else printf("No\n");
continue;
}
if(n%2==1)
{
printf("No\n");
continue;
}
else
{
int x=0;
int y=0;
for(int i=0;i<n;i++)
{
if(a[i]=='(')x++;
else y++;
}
if(x!=y)
{
printf("No\n");
continue;
}
int sum=0;
int flag=0;
for(int i=0;i<n;i++)
{
if(a[i]=='(')
{
sum++;
}
else
{
if(sum==0)flag++;
else sum--;
}
}
if(flag==0)
{
printf("Yes\n");
}
else if(flag==1||flag==2)
{
printf("Yes\n");
}
else
{
printf("No\n");
}
}
}
return 0;
}