HDU 5831 Rikka with Parenthesis II (贪心)

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

题解：

若左括号和右括号数量不等，输出’No’

若n==2，这个串刚好是’()’，因为必须交换的原因，也是’No’。

其它情况：

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.

若大于等于三队，则是’No’。（采用栈的方法判断括号的匹配情况）

#include <iostream>
#include<cstdio>
#include<algorithm>
#include<stack>
#include<cstring>

using namespace std;
int T,n,l,r;
char ch[];
stack<char>s;
int main()
{
    scanf("%d",&T);
    for(;T>;T--)
    {
        scanf("%d",&n);
        scanf("%s",&ch);
        if (n== && ch[]=='(' && ch[]==')')
        {
            printf("No\n");
            continue;
        }
        while(!s.empty()) s.pop();
        l=; r=;
        for(int i=;i<n;i++)
        {
            if (ch[i]=='(') s.push(ch[i]);
             else
             {
                 if (!s.empty()) s.pop();
                  else l++;
             }
        }
        r=s.size();
        if (r!=l) {printf("No\n"); continue;}
        if (l>) {printf("No\n"); continue;}
        printf("Yes\n");
    }
    return ;
}