# UVa 673 - Parentheses Balance
## Online Judge

## 解題思路
蠻簡單的一題，23和33行加入counter，是因為已知道答案，所以跳出for loop。
p.s.少了那邊的判斷會多60ms
## 解題中出現的bug
沒什麼大問題。
## Code

#include <iostream>
#include <string>
#include <stack>
using namespace std;

int main()
{
	int test_case;
	cin >> test_case;
	getchar();
	while(test_case--)
	{
		bool counter = true;
		string token;
		stack<char> Stack;
		getline(cin, token);
		for (size_t i = 0; i < token.size()&&counter==true; i++)
		{
			if (token[i] == ')'&&!Stack.empty())
			{
				if (Stack.top() == '(')
					Stack.pop();
				else //keep quickly break this for loop
				{
					counter = false;
					break;
				}
			}
			else if (token[i] == ']'&& !Stack.empty())
			{
				if(Stack.top() == '[')
					Stack.pop();
				else
				{
					counter = false;
					break;
				}
			}
			else
				Stack.push(token[i]);
		}
		if (Stack.empty())
			cout << "Yes" << endl;
		else
			cout << "No" << endl;
	}
	return 0;
}

###### tags: `UVA code` `cpp` `林基成-C++` `Awwwolf的刷題之路`