Parentheses Balance

Description:
You are given a string consisting of parentheses () and []. A string of this type is said to be correct:
(a) if it is the empty string
(b) if A and B are correct, AB is correct,
© if A is correct, (A) and [A] is correct.
Write a program that takes a sequence of strings of this type and check their correctness. Your program can assume that the maximum string length is 128.
Input
The file contains a positive integer n and a sequence of n strings of parentheses ‘()’ and ‘[]’, one string
a line.
Output
A sequence of ‘Yes’ or ‘No’ on the output file.
Sample Input
3
([])
(([()])))
([()])()
Sample Output
Yes
No
Yes

利用STL中的栈比较容易解决。
stackname;声明一个T类型的名称为name的栈。
注意因为空串应该输出Yes，所以不可以用string读入表达式，它会跳过空格，换行符。这里利用的是gets来读入，通过判断读入的字符串是否等于换行符来解决空串。此外，我用的是cin>>n；所以要用getchar()将回车滤掉，要不然输入n后换行就会输出Yes。

表达式中如果是左括号则进栈，若是右括号则取出栈顶元素判断是否和这个右括号匹配，注意！先判断栈是否是非空的才可以进行取元素操作。

#include<iostream>
#include<stack>
#include<cstring>
#include<string>
using namespace std;
stack<char>exp;
int main(){
	int n;
	cin>>n;
	getchar();//滤掉一个空格 
	char s[200];
	while(n--){
		int flag=1;
		gets(s);//判断是否是空串，不能用string了 
		if(strcmp(s,"\n")==0){
			cout<<"Yes"<<endl;
			continue;
		}
//		int len=s.length();
		for(int i=0;s[i];i++){
			if(s[i]=='['||s[i]=='(')
				exp.push(s[i]);
			else if(s[i]==']'){
				if(exp.empty()){
					flag=0;
					break;
				}
				char ch=exp.top();
				if(ch=='[')
					exp.pop();
			}
			else if(s[i]==')'){
				if(exp.empty()){
					flag=0;
					break;
				}
				char ch=exp.top();
				if(ch=='(')
					exp.pop();
			}
		}
		if(!exp.empty())
			flag=0;
		if(flag)
			cout<<"Yes"<<endl;
		else
			cout<<"No"<<endl;
		while(!exp.empty())
			exp.pop();
	}
	return 0;
}