HDU 5831 Rikka with Parenthesis II（栈的使用）

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 78    Accepted Submission(s): 61

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.
 

 

Author

学军中学

 

Source

2016 Multi-University Training Contest 8 

 

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总结：

看到这道题不难想到以前写过的括号匹配问题，但是有一点不同的是可以通过一次改变来修正括号，遇到没思路的题不如先看一看样例，把自己解决的过程慢慢的搞清楚，也许问题就得到解决了，在手工推倒的过程中发现只要是交换位置那么一定是把前方的）和后方的（进行交换，之后又发现除去已经匹配好的括号之后最外层的括号一定是）。。（这种形式，那么我们只需要交换这两个括号就可以了

////  main.cpp//  Rikka with Parenthesis II////  Created by 张嘉韬 on 16/8/11.//  Copyright © 2016年 张嘉韬. All rights reserved.//#include <iostream>#include <cstdio>#include <stack>using namespace std;const int maxn=100000+10;int main(int argc, const char * argv[]) {    int T;    scanf("%d",&T);    while(T--)    {        int n,counter1=0,counter2=0;        stack <char> st;        scanf("%d",&n);        if(n==0) {printf("Yes\n"); continue;}        char s[maxn],left[maxn];        for(int i=1;i<=n;i++)        {            cin>>s[i];            if(s[i]=='(') counter1++;            else counter2++;        }        if(counter1!=counter2) {printf("No\n"); continue;}        if(n==2&&s[1]=='('&&s[2]==')') {printf("No\n"); continue;}        for(int i=1;i<=n;i++)        {            if(s[i]=='(')            {                st.push(s[i]);            }            else            {                if(st.empty()==1) st.push(s[i]);                else st.pop();            }        }        if(st.empty()==1) {printf("Yes\n"); continue;}        int p=0;        while(st.empty()!=1)        {            left[++p]=st.top();            st.pop();        }        left[1]='(';        left[p]=')';        for(int i=1;i<=p;i++)        {            if(left[i]=='(')            {                st.push(left[i]);            }            else            {                if(st.empty()==1) st.push(left[i]);                else st.pop();            }        }        if(st.empty()==1) {printf("Yes\n"); continue;}        else {printf("No\n"); continue;}    }    return 0;}