[class2] (백준 4949) 균형잡힌 세상

문제

세계는 균형이 잘 잡혀있어야 한다. 양과 음, 빛과 어둠 그리고 왼쪽 괄호와 오른쪽 괄호처럼 말이다.

정민이의 임무는 어떤 문자열이 주어졌을 때, 괄호들의 균형이 잘 맞춰져 있는지 판단하는 프로그램을 짜는 것이다.

문자열에 포함되는 괄호는 소괄호("()") 와 대괄호("[]")로 2종류이고, 문자열이 균형을 이루는 조건은 아래와 같다.

• 모든 왼쪽 소괄호("(")는 오른쪽 소괄호(")")와만 짝을 이뤄야 한다.
• 모든 왼쪽 대괄호("[")는 오른쪽 대괄호("]")와만 짝을 이뤄야 한다.
• 모든 오른쪽 괄호들은 자신과 짝을 이룰 수 있는 왼쪽 괄호가 존재한다.
• 모든 괄호들의 짝은 1:1 매칭만 가능하다. 즉, 괄호 하나가 둘 이상의 괄호와 짝지어지지 않는다.
• 짝을 이루는 두 괄호가 있을 때, 그 사이에 있는 문자열도 균형이 잡혀야 한다.

정민이를 도와 문자열이 주어졌을 때 균형잡힌 문자열인지 아닌지를 판단해보자.

입력

하나 또는 여러줄에 걸쳐서 문자열이 주어진다. 각 문자열은 영문 알파벳, 공백, 소괄호("( )") 대괄호("[ ]")등으로 이루어져 있으며, 길이는 100글자보다 작거나 같다.

입력의 종료조건으로 맨 마지막에 점 하나(".")가 들어온다.

출력

각 줄마다 해당 문자열이 균형을 이루고 있으면 "yes"를, 아니면 "no"를 출력한다.

제한

예제 입력 1

So when I die (the [first] I will see in (heaven) is a score list).
[ first in ] ( first out ).
Half Moon tonight (At least it is better than no Moon at all].
A rope may form )( a trail in a maze.
Help( I[m being held prisoner in a fortune cookie factory)].
([ (([( [ ] ) ( ) (( ))] )) ]).
 .
.

예제 출력 1

yes
yes
no
no
no
yes
yes

힌트

7번째의 " ."와 같이 괄호가 하나도 없는 경우도 균형잡힌 문자열로 간주할 수 있다.

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

출처

ICPC > Regionals > Asia Pacific > Japan > Japan Domestic Contest > 2011 Japan Domestic Contest B번

• 문제의 오타를 찾은 사람: chminoo
• 잘못된 번역을 찾은 사람: jh05013
• 문제를 번역한 사람: occidere

 

#include <iostream>
#include <stack>
#include <cstring>
using namespace std;
int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	char input[102];
	stack<char> stringDeck;
	while (1)
	{
		cin.getline(input, 102 ,'.');
		cin.ignore();
		if (input[0] == '\0')
			break;
		for (int i = 0; i < 102; i++)
		{
			if (input[i] == '\0')
				break;
			if (input[i] == '(' || input[i] == '[')
			{
				stringDeck.push(input[i]);
			}
			else if (input[i] == ')')
			{
				if (!stringDeck.empty() && stringDeck.top() == '(')
					stringDeck.pop();
				else
					stringDeck.push(input[i]);
			}
			else if (input[i] == ']')
			{
				if (!stringDeck.empty() && stringDeck.top() == '[')
					stringDeck.pop();
				else
					stringDeck.push(input[i]);
			}
		}
		if (stringDeck.empty())
			cout << "yes\n";
		else
		{
			cout << "no\n";
			while (!stringDeck.empty()) stringDeck.pop();
		}
	}
	return 0;
}