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);
    while(n--){
        gets(c);//因为可能读入空串，所以用gets 
        if(strcmp(c,"
")==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
");
        else printf("Yes
");
    }
    return 0;
}