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[500];
int n;
int pd(char q,char e){//匹配
if(q=='(' && e==')')return 1;
if(q=='[' && e==']')return 1;
return 0;
}
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")==0){//空串合法
printf("Yes");
continue;
}
int flag=0;
stack<char>st;
int L=strlen(c);
for(int i=0;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=1;
break;
}
if(pd(st.top(),c[i])==1) st.pop();
else{
flag=1;
break;
}
}
}
if(flag==1 || !st.empty())printf("No\n");
else printf("Yes\n");
}
return 0;
}