UVa 673 Parentheses Balance (栈)

673 - Parentheses Balance

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=614

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,

(c)

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      

技巧：在栈底加一个元素，减少代码量。

完整代码：

/*0.029s*/

#include<cstdio>
#include<cstring>
#include<stack>
using namespace std;

stack<char> s;
char str[130];

int main(void)
{
	int t;
	scanf("%d", &t);
	getchar();
	while (t--)
	{
		gets(str);
		int len = strlen(str);
		if (len & 1)
			puts("No");
		else
		{
			if (!s.empty())
				s.pop();
			s.push('0');///“记号”
			for (int i = 0; i < len; ++i)
			{
				if (str[i] == '(' || str[i] == '[')
					s.push(str[i]);
				else if (str[i] == ')')
				{
					if (s.top() == '(')
						s.pop();
					else
					{
						s.push('1');
						break;
					}
				}
				else/// ']'
				{
					if (s.top() == '[')
						s.pop();
					else
					{
						s.push('1');
						break;
					}
				}
			}
			puts(s.top() == '0' ? "Yes" : "No");
		}
	}
	return 0;
}