HDU5831(2016多校第八场)———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): 16    Accepted Submission(s): 16

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.

#include <cstdio>#include <cstring>#include <string>#include <iostream>#include <algorithm>#include <vector>#include <queue>#include <set>#include <map>#include <string>#include <cmath>#include <cstdlib>using namespace std;const int MAXN =100000+10;const int INF =1000000007;char a[MAXN];int main(){    //freopen("in.txt","r+",stdin);    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        getchar();        scanf("%s",a);        int res=0;        if(n%2==1){            printf("No\n");            continue;        }        if(n==2){            if(a[0]=='('&&a[1]==')'){                printf("No\n");            }            else{                printf("Yes\n");            }            continue;        }        int ok=1;        int flag=1;        for(int i=0;i<n;i++){            if(a[i]=='(')                res++;            else                res--;            if(res<0&&flag){                for(int j=n-1;j>i;j--){                    if(a[j]=='('){                        swap(a[i],a[j]);                        res=1;                        flag=0;                        break;                    }                }            }            else if(flag==0&&res<0)            {                ok=0;                break;            }        }        if(!ok){            printf("No\n");            continue;            }        if(res!=0){            printf("No\n");            continue;        }        printf("Yes\n");    }}