Problem Description
As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of student want to get back to school by train(because the trains in the Ignatius Train Station is the fastest all over the world ^v^). But here comes a problem, there is only one
railway where all the trains stop. So all the trains come in from one side and get out from the other side. For this problem, if train A gets into the railway first, and then train B gets into the railway before train A leaves, train A can't leave until train
B leaves. The pictures below figure out the problem. Now the problem for you is, there are at most 9 trains in the station, all the trains has an ID(numbered from 1 to n), the trains get into the railway in an order O1, your task is to determine whether the
trains can get out in an order O2.

 
Input
The input contains several test cases. Each test case consists of an integer, the number of trains, and two strings, the order of the trains come in:O1, and the order of the trains leave:O2. The input is terminated by the end of file. More details in the Sample
Input.
 
Output
The output contains a string "No." if you can't exchange O2 to O1, or you should output a line contains "Yes.", and then output your way in exchanging the order(you should output "in" for a train getting into the railway, and "out" for a train getting out of
the railway). Print a line contains "FINISH" after each test case. More details in the Sample Output.
 
Sample Input
3 123 321
3 123 312
 
Sample Output
Yes.
in
in
in
out
out
out
FINISH
No.
FINISH
Hint
Hint
For the first Sample Input, we let train 1 get in, then train 2 and train 3.
So now train 3 is at the top of the railway, so train 3 can leave first, then train 2 and train 1.
In the second Sample input, we should let train 3 leave first, so we have to let train 1 get in, then train 2 and train 3.
Now we can let train 3 leave.
But after that we can't let train 1 leave before train 2, because train 2 is at the top of the railway at the moment.
So we output "No.".
 

栈的简单应用,本题具体解答在白书上有.

以下附上代码:用c++交会错。用G++就不会。。

。求解释

#include<iostream>
#include<stack>
#include<cstring>
#include<cstdio>
using namespace std;
int a[10000],ans[20000],b[10000];
stack<int>ss;
int main()
{
int n,i,j,k;
string s1,s2;
while(scanf("%d",&n)!=EOF){
cin>>s1>>s2;
for(i=1;i<=n;i++) a[i]=s1[i-1]-'0';
for(j=1;j<=n;j++) b[j]=s2[j-1]-'0';
while(!ss.empty()) ss.pop();
k=0,j=1;
for(i=1;i<=n;i++) {
ss.push(a[i]);
ans[++k]=1;
while(!ss.empty() && ss.top()==b[j]) {
ss.pop();
ans[++k]=2;
j++;
}
}
if(k==2*n) {
printf("Yes.\n");
for(i=1;i<=2*n;i++) {
if(ans[i]==1) printf("in\n");
else printf("out\n");
}
}
else printf("No.\n");
printf("FINISH\n");
}
return 0;
}

HDU Train Problem I (STL_栈)的更多相关文章

  1. HDU Train Problem I 1022 栈模拟

    题目大意: 给你一个n 代表有n列 火车,  第一个给你的一个字符串 代表即将进入到轨道上火车的编号顺序, 第二个字符串代表的是 火车出来之后到顺序, 分析一下就知道这,这个问题就是栈, 先进后出吗, ...

  2. HDU 1022 Train Problem I(栈的操作规则)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1022 Train Problem I Time Limit: 2000/1000 MS (Java/Ot ...

  3. HDU 1022.Train Problem I【栈的应用】【8月19】

    Train Problem I Problem Description As the new term comes, the Ignatius Train Station is very busy n ...

  4. hdu 1022 Train Problem I(栈的应用+STL)

    Train Problem I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) ...

  5. Train Problem I--hdu1022(栈)

    Train Problem I Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  6. HDU 1022 Train Problem I(栈模拟)

    传送门 Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot of st ...

  7. HDOJ/HDU 1022 Train Problem I(模拟栈)

    Problem Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot o ...

  8. hdu Train Problem I(栈的简单应用)

    Problem Description As the new term comes, the Ignatius Train Station is very busy nowadays. A lot o ...

  9. hdu 1022 Train Problem I(栈)

    #include<iostream> #include<vector> #include<cstring> #include<string> #incl ...

随机推荐

  1. 【剑指Offer学习】【面试题63:二叉搜索树的第k个结点】

    题目:给定一棵二叉搜索树,请找出当中的第k大的结点. 解题思路 假设依照中序遍历的顺序遍历一棵二叉搜索树,遍历序列的数值是递增排序的. 仅仅须要用中序遍历算法遍历一棵二叉搜索树.就非常easy找出它的 ...

  2. [Ionic] Align and Size Text with Ionic CSS Utilities

    The Ionic framework provides several built-in CSS Utilities or directives that you can leverage when ...

  3. 嵌入式linux和pc机的linux对照

    linux本身具备的非常大长处就是稳定,内核精悍,执行时须要的资源少.嵌入式linux和普通linux并无本质差别. 在嵌入式系统上执行linux的一个缺点就是其核心架构没有又一次设计过,而是直接从桌 ...

  4. 学习笔记一:关于directx sdk的安装于一些概念

    关于directx sdk开发环境的安装: 在百度搜索了directx sdk,进入了微软的官网,下载了DXSDK_Jun10.exe 百度网盘:http://pan.baidu.com/s/1o6r ...

  5. 2016.02.25,英语,《Vocabulary Builder》Unit 02

    ag:来自拉丁语do.go.lead.drive,an agenda是要做事情的清单,an agent是代表他们做事的人,同时也是为他人做事的机构.拉丁语litigare包括词根lit,即lawsui ...

  6. Oracle 常见的33个等待事件

    一. 等待事件的相关知识: 1.1 等待事件主要可以分为两类,即空闲(IDLE)等待事件和非空闲(NON-IDLE)等待事件. 1). 空闲等待事件指Oracle正等待某种工作,在诊断和优化数据库的时 ...

  7. Java类的根Object

    一.Object类介绍 Object全名java.lang.Object,java.lang包在使用的时候无需显示导入,编译时由编译器自动导入.Object类是类层次结构的根,Java中所有的类从根本 ...

  8. linux服务器网站安全狗安装教程

    1.下载服务器安全狗和服务器网站安全狗,选择好版本.http://download.safedog.cn/safedog_linux64.tar.gz 这个是网站安全狗的下载地址2.登录centos进 ...

  9. Selenium键盘鼠标操作总结

    鼠标操作 org.openqa.selenium.interactions.Actions 1.给元素设置焦点. 有时对于a标签等,为了不跳转到别的链接,但是需要设置焦点时就可使用. action.m ...

  10. spring事务,TransactionAspectSupport.currentTransactionStatus().setRollbackOnly();

    在aop配置事务控制或注解式控制事务中,try...catch...会使事务失效,可在catch中抛出运行时异常throw new RuntimeException(e)或者手动回滚Transacti ...