UVa 673 - Parentheses Balance

Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld
& %llu
Description

Parentheses Balance

You are given a string consisting of parentheses () and []. A string of this type is said to becorrect:

(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：

输入一个只含有 ‘(’ 、‘)’ 、‘[’ 、‘]’ 的序列，判断并输出括号是否匹配。

题目解析：

如果遇到 '(' 、'[' 就入栈，如果遇到 ')' 、']' 先判断栈是否为空，再判度栈顶是否匹配，如果匹配就弹出，否则就入栈。

最后判断是否为空，如果为空就输出Yes，否则No

#include <iostream>
#include <stdio.h>
#include <stack>
#include <string.h>
using namespace std;
const int N = 150;
char str
;
int main() {
	int t;
	while(scanf("%d",&t) != EOF) {
		getchar();
		while(t--) {
			gets(str);
			//puts(str);

			stack<char> st;
			int len = strlen(str);
			for(int i = 0; i < len; i++) {
				if(str[i] == '(' || str[i] == '[') {
					st.push(str[i]);
				}
				else {
					if(st.size() && str[i] == ')' && st.top() == '(') {
						st.pop();
					}
					else
						if( st.size() && str[i] == ']' && st.top() == '[') {
							st.pop();
						}
						else
							st.push(str[i]);
				}
			}

			if(st.size() == 0) {
				printf("Yes\n");
			}
			else {
				printf("No\n");
			}
		}
	}
	return 0;
}