HDU 5831 Rikka with Parenthesis II 【贪心】

Rikka with Parenthesis II

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1612    Accepted Submission(s): 709

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

 

题解：左括号和右括号个数不相等是No，n=2的时候如果是()也是no，因为必须交换一次，就会变成)(，所以是No。否则如果出现一个没有与其相匹配的右括号，就是右括号出现在与他匹配的左括号之前，如果这种情况出现了三次或三次以上就是No，其余是Yes。

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<queue>#include<stack>#include<vector>#include<map>#include<set>#include<algorithm>using namespace std;#define ll long long#define ms(a,b)  memset(a,b,sizeof(a))const int M=1e6+10;const int inf=0x3f3f3f3f;const int mod=1e9+7;int i,j,k,n,m;int sg[M];int vis[M];int gcd,x,y;ll p,q;ll ans;ll sum[M];int a[M];char s[M];int main(){    int t;    scanf("%d",&t);    while(t--){        scanf("%d",&n);        scanf("%s",s);        ms(a,0);        int sum=0;        int sum1=0;        int sum2=0;        int ans=0;        int len=strlen(s);        for(int i=0;i<n;i++){            if(s[i]=='('){                a[i]=1,sum1++;                sum++;            }            else {                a[i]=-1,sum2++;                if(sum>0)sum--;                else ans++;            }        }        if(sum1!=sum2)printf("No\n");        else {            if(ans>=3)printf("No\n");            else if(len==2&&ans==0)printf("No\n");            else printf("Yes\n");        }    }    return 0;}