Rikka with Parenthesis II
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 0 Accepted Submission(s): 0
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.
题目大意:
给你n长度的一个括号串,问你是否能够在必须交换两个括号的情况下,使得最终交换得到的括号串是匹配的
题解:
若左括号和右括号数量不等,输出’No’
若n==2,这个串刚好是’()’,因为必须交换的原因,也是’No’。
其它情况:
①0个需要挪动的括号,而且n>=4,那么随便挪动一对括号就行。
②1个需要挪动的括号,那么一定有一个右括号也需要挪动与之匹配:)(
③2个需要挪动的括号,那么一定有两个右括号也需要挪动与之匹配:))((挪动14就可以达到目的.
若大于等于三队,则是’No’。(采用栈的方法判断括号的匹配情况)
#include <iostream> #include<cstdio> #include<algorithm> #include<stack> #include<cstring> using namespace std; int T,n,l,r; char ch[100005]; stack<char>s; int main() { scanf("%d",&T); for(;T>0;T--) { scanf("%d",&n); scanf("%s",&ch); if (n==2 && ch[0]=='(' && ch[1]==')') { printf("No "); continue; } while(!s.empty()) s.pop(); l=0; r=0; for(int i=0;i<n;i++) { if (ch[i]=='(') s.push(ch[i]); else { if (!s.empty()) s.pop(); else l++; } } r=s.size(); if (r!=l) {printf("No "); continue;} if (l>2) {printf("No "); continue;} printf("Yes "); } return 0; }