首页 > 代码库 > UVA 637 Parentheses Balance（栈）
UVA 637 Parentheses Balance（栈）
题目代号：HDU 1237
题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=103&page=show_problem&problem=614
题目原文：
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：输入一个包含()和[]的括号序列，判断是否合法。1.空串合法；2.如果A和B都合法，则AB合法；3.如果A合法则(A)和[A]都合法。
栈的练习题
通过代码：
# include <stdio.h>
# include <string.h>
# include <stdlib.h>
# include <iostream>
# include <fstream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <math.h>
# include <algorithm>
using namespace std;
# define pi acos(-1.0)
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define For(i,n,a) for(int i=n; i>=a; --i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define Fo(i,n,a) for(int i=n; i>a ;--i)
typedef long long LL;
typedef unsigned long long ULL;
int main()
{
int a,b,c,d,t;
char ch,zhan[200];
cin>>t;
getchar();
while(t--)
{
int top=-1;
bool flag=true;
while((ch=getchar())!=‘\n‘)
{
if(ch==‘(‘)
{
zhan[++top]=‘(‘;
}
else if(ch==‘[‘)
{
zhan[++top]=‘[‘;
}
else if(ch==‘)‘)
{
if(zhan[top]==‘(‘)
{
top--;
}
else flag=false;
}
else if(ch==‘]‘)
{
if(zhan[top]==‘[‘)
{
top--;
}
else flag=false;
}
}
if(top>=0)flag=false;
if(flag)cout<<"Yes"<<endl;
else cout<<"No"<<endl;
}
return 0;
}
UVA 637 Parentheses Balance（栈）
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。