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 

YesNoYes

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;}