UVA673 Parentheses Balance



Parentheses Balance

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

这一个简答的题竟然提交了五遍真是醉了。

错误的原因

第一空字符串输出yes,注意到这个了，也有判断条件，还是不对，因为输入用的scanf，当输入空行是没反应，最后换成了gets

第二第二个样例老是意外中断，一直没发现问题，最后意识到是当')'或者‘】'多的时候，栈已经空了，此时取不到栈顶元素，所以意外退出

第三自己构造了一些样例不对，最后发现是在每一次的输入前没有清空栈

就是这些问题，

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

stack<char> Stack;
int main()
{
    int t;
    scanf("%d",&t);
    getchar();
    while(t--)
    {
        char s[150];
        gets(s);
        int len=strlen(s);
        if(len==0) printf("Yes\n");
        else
        {
            int flag=0;
            while(Stack.size())
                Stack.pop();
            for(int i=0;i<len;i++)
            {
                char ch;
                if(s[i]=='(' || s[i]=='[')
                    Stack.push(s[i]);
                if(s[i]==')')
                {
                    if(Stack.size()==0)
                    {
                        flag=1;
                        break;
                    }
                    else
                    {
                        ch=Stack.top();
                        if(ch=='(')
                        Stack.pop();
                        else
                        {
                           flag=1;
                           break;
                        }
                    }

                 }
                 if(s[i]==']')
                 {
                     if(Stack.size()==0)
                     {
                        flag=1;
                        break;
                     }
                     else
                     {
                         ch=Stack.top();
                         if(ch=='[')
                         Stack.pop();
                         else
                         {
                           flag=1;
                           break;
                         }
                     }
                  }
               }

            if(!flag && Stack.size()==0)
                printf("Yes\n");
            else printf("No\n");
        }
    }
    return 0;
}