UVa 673 Parentheses Balance -SilverN

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

用栈模拟匹配括号

 /*UVa 673 Parentheses Balance*/
 #include<iostream>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<stack>
 using namespace std;
 char c[];
 int n;
 int pd(char q,char e){//匹配
     if(q=='(' && e==')')return ;
     if(q=='[' && e==']')return ;
     return ;
 }
 int main(){
     scanf("%d\n",&n);
     while(n--){
         gets(c);//因为可能读入空串，所以用gets
         if(strcmp(c,"\n")==){//空串合法
             printf("Yes");
             continue;
         }
         int flag=;
         stack<char>st;
         int L=strlen(c);
         for(int i=;i<L;i++){
             if(c[i]=='(' || c[i]=='[') st.push(c[i]);//左括号入栈
             else{//右括号处理
                 if(st.empty()){
                     flag=;
                     break;
                 }
                 if(pd(st.top(),c[i])==) st.pop();
                 else{
                     flag=;
                     break;
                     }
             }
         }
         if(flag== || !st.empty())printf("No\n");
         else printf("Yes\n");
     }
     return ;
 }