HDU 5831 Rikka with Parenthesis II

题目描述：

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.

题目分析：

HDU多校第八场1011。
给一个n个字符的串，串由’(‘和’)’组成，其中你必须进行操作，（注意是必须），将这个串中的两个括号互相交换。
进行分析之后可以看到，若左括号和右括号数量不等，输出’No’,或者这个串刚好是’()’，因为必须交换的原因，也是’No’。其它情况，这个括号序列中若有一对不合法，那么交换这一对就可以。若有两对不合法，也就是类似’))((‘类型的，我们可以将其中最左的和最右的交换，就变成’()()’的合法类型了。因此我们可以推断，除去其他情况，这个串中若有一对或者两对不符合括号，输出’Yes’，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）

代码如下：

#include <iostream>#include <cstdio>#include <cstring>#include <stack>using namespace std;int T;stack<int>sta;int l,r;int n;char s[100010];int main(){    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        scanf("%s",s);        if (n==2 && s[0]=='(' && s[1]==')') {printf("No\n");continue;}        l=0;        r=0;        while(!sta.empty()) sta.pop();        for(int i=0; i<n; i++)        {            if (s[i]=='(') sta.push(i);            else if (s[i]==')')            {                if (sta.empty()) r++;                else sta.pop();            }        }        l=sta.size();        if (l!=r) {printf("No\n");continue;}        if (l>2) {printf("No\n");continue;}        printf("Yes\n");    }    return 0;}