HDU 5831 乱搞

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.


题意：给你一串括号，交换一次任意两个括号的位置，能不能使括号完全匹配。

题解：先消去所有能匹配的，如果没有匹配完全，那就选取最左边的')'和最右边的'('交换即可，如果匹配完全，那就随便选取两个相同的括号交换即可，所以这里也要特判n为2的情况，即（）。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n;
char str[100005];
char q[100005];
int main(){
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        scanf("%s",str);
        int sum1=0,sum2=0;
        if(n%2){
            printf("No\n");
        }
        else{
            int qend=0;
            for(int i=1;i<=n;i++){
                if(str[i-1]=='('){
					sum1++;
					q[++qend]='(';
                }
                else{
					sum2++;
					if(q[qend]=='('){
						qend--;
					}
					else q[++qend]=')';
                }
            }
            if(sum1!=sum2){
                printf("No\n");
                continue;
            }
            if(!qend){
                if(strlen(str)==2)printf("No\n");
                else printf("Yes\n");
                continue;
            }
            int i,lef=-1,rig=-1;
            for(i=1;i<=qend;i++){
                if(q[i]==')'){
					lef=i;
					break;
                }
            }
            for(i=qend;i>=1;i--){
                if(q[i]=='('){
					rig=i;
					break;
                }
            }
            swap(q[lef],q[rig]);
            while(qend){
                if(q[qend]==')'&&q[qend-1]=='('){
                    qend-=2;
                }
                else break;
            }
            if(qend==0)printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}