UVaOJ673---Parentheses Balance

最新推荐文章于 2017-12-26 16:08:50 发布

原创最新推荐文章于 2017-12-26 16:08:50 发布 · 820 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#CC++ #算法 #ACM

===ACM|ICPC=== 同时被 2 个专栏收录

108 篇文章

订阅专栏

Volume1-数据结构基础

30 篇文章

订阅专栏

本文介绍了一个简单的算法，用于检查由括号组成的字符串是否正确配对。该算法利用栈数据结构来跟踪并验证每一对括号的正确性。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

673 - Parentheses Balance

Time limit: 3.000 seconds

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

Miguel Revilla
2000-08-14

水题~~~~~~~~~~~~~~~~~~~~

#include <iostream>
#include <stack>
#include <string>
using namespace std;
int n;
string str;
enum Parentheses
{
   BRACKET,
   PARENTHESIS,
};
stack<Parentheses> parentheses;
bool isYes;
int main()
{
    cin>>n;
    cin.get();
    while (n --)
    {
       while (! parentheses.empty())
       {
          parentheses.pop();
       }
       getline(cin, str);
       isYes = true;
       for (unsigned int i = 0; i < str.length(); i++)
       {
          if (str[i] == '(')
          {
             parentheses.push(PARENTHESIS);
          }
          else if (str[i] == '[')
          {
             parentheses.push(BRACKET);
          }
          else if (str[i] == ')')
          {
             if (parentheses.empty() || parentheses.top() != PARENTHESIS)
             {
                isYes = false;
                break;
             }
             else
             {
                parentheses.pop();
             }
          }
          else if (str[i] == ']')
          {
             if (parentheses.empty() || parentheses.top() != BRACKET)
             {
                isYes = false;
                break;
             }
             else
             {
                parentheses.pop();
             }
          }
       }
       if (! parentheses.empty())
       {
          isYes = false;
       }
       if (isYes)
       {
          cout<<"Yes"<<endl;
       }
       else
       {
          cout<<"No"<<endl;
       }
    }
    return 0;
}