Rikka with Parenthesis II

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5831

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 $S_i,S_j$.
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(1100
For each testcase, the first line contains an integers n(1

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.

Source

2016 Multi-University Training Contest 8


##题意：

括号匹配问题：空、XY、(X) 为合法字串.
给出一个字串，是否可以经过且只经过一次交换操作，使得结果串合法.
(不能不交换，不能与自己位置交换)


##题解：

考虑交换操作：(必须换且只能换一次).
①. 如果原串本身就合法，长度为2时："()"->"No", ")("->"Yes".
长度大于2时一定为"Yes", 因为可以直接交换两个相同的括号.
②. 如果原串非法，那么交换时一定交换的不同的符号(否则没用).
那么符合条件的串一定是把一个 '(' -> ')' 且一个 ')' -> '(' . 那么只考虑变换即可.

考虑如何判断一个串是否合法的过程：
依次处理字符，若是'('则入栈，若是')'则从栈中弹出一个'('. 若没有'('则不合法.
那么此题就是上述过程的变种，在处理过程中允许两次变换. 由于'('->')'的时机不方便考虑, 这里就只考虑')'->'('.
①. 如果当前是'('，直接入栈.
②. 如果当前是')'，如果栈非空，则弹出一个'('; 如果栈空就把当前的')'变成'('入栈. (标记最多只能变化一次).

用flag标记是否有将')'变为'('的操作. 结果栈要么为空，要么全是'('.
1. 如果整个字串没有被处理完，那么肯定是"No".
2. 如果flag=0, 那么要求没有'('剩下.
3. 如果flag=1, 那么结果栈中的'('只能是两个. "((" -> "()".

官方题解：
最优情况下一定交换第一个右括号和最后一个左括号，交换后判断一下即可。 时间复杂度 O(n).


##代码：
``` cpp
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define LL long long
#define eps 1e-8
#define maxn 101000
#define mod 100000007
#define inf 0x3f3f3f3f
#define mid(a,b) ((a+b)>>1)
#define IN freopen("in.txt","r",stdin);
using namespace std;

char str[maxn];

stack s;

int main(int argc, char const *argv[])

{

//IN;

int t; cin >> t;

while(t--)

{

    int n; scanf("%d", &n);

    while(!s.empty()) s.pop();

    scanf("%s", str);

    if(n == 2) {

        if(str[0]=='(' && str[1]==')') {

            puts("No");

            continue;

        }

    }

    int i;

    int flag1 = 0;

    for(i=0; i<n; i++) {

        if(str[i] == '(') {

            s.push('(');

        } else {

            if(!s.empty()) s.pop();

            else {

                if(flag1) break;

                flag1 = 1;

                s.push('(');

            }

        }

    }

    if(i == n) {

        if(!flag1) {

            if(s.empty()) puts("Yes");

            else puts("No");

        }

        else {

            if(s.size() != 2) puts("No");

            else puts("Yes");

        }

    }

    else puts("No");

}

return 0;

}