UVA 673 栈 括号匹配

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

 题目大意和分析：

括号匹配。从头遍历。若输入的是“（”，“[”。将他们放入栈中。如果是“）”，或“]”。则判断栈是否为空。不为空。则看栈顶的元素。如果是“（”或“[”。则将栈顶元素弹出。

否则将tmp 置为 1.最后判断栈是否为空，为空切他们tmp = 0，则输出Yes。反之输出No。

这是一道栈的经典题，但也是一道水题，进栈出栈就如此简单的操作，无复杂算法。

题目链接：http://acm.uva.es/local/online_judge/search_uva.html

AC代码：

#include <iostream>
#include <string>
#include <cstring>
#include <cstdio>
#include <stack>
#include <vector>
using namespace std;
const int MAX=128+1;

int n,len;
string str;//读入
int main()
{
    cin >> n;
    cin.get ( );//控制回车
    while ( n-- )
    {
        bool tmp = 0;
        stack < char > sta;//创建一个栈
        getline ( cin, str );//将读到的放入str中
        int len = str.length ( );
        for ( int i = 0; i < len; ++i ) 
        {
            if ( str[i] == '(' || str[i] == '[' )
                sta.push( str[i] );
            else if ( str[i] == ')' && !sta.empty( ) && sta.top ( ) == '(' ) 
                sta.pop ( );
            else if ( !sta.empty ( ) && str[i] == ']' && sta.top ( ) == '[' ) 
                sta.pop ( );
            else tmp = 1;
        }
        if(tmp==0 && sta.empty())
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    return 0;
}