Elven Postman
Elven Postman
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1108 Accepted Submission(s): 587
So, as a elven postman, it is crucial to understand how to deliver the mail to the correct room of the tree. The elven tree always branches into no more than two paths upon intersection, either in the east direction or the west. It coincidentally looks awfully like a binary tree we human computer scientist know. Not only that, when numbering the rooms, they always number the room number from the east-most position to the west. For rooms in the east are usually more preferable and more expensive due to they having the privilege to see the sunrise, which matters a lot in elven culture.
Anyways, the elves usually wrote down all the rooms in a sequence at the root of the tree so that the postman may know how to deliver the mail. The sequence is written as follows, it will go straight to visit the east-most room and write down every room it encountered along the way. After the first room is reached, it will then go to the next unvisited east-most room, writing down every unvisited room on the way as well until all rooms are visited.
Your task is to determine how to reach a certain room given the sequence written on the root.
For instance, the sequence 2, 1, 4, 3 would be written on the root of the following tree.
For each test case, there is a number n(n≤1000) on a line representing the number of rooms in this tree. n integers representing the sequence written at the root follow, respectively a1,...,an where a1,...,an∈{1,...,n}.
On the next line, there is a number q representing the number of mails to be sent. After that, there will be q integers x1,...,xq indicating the destination room number of each mail.
Note that for simplicity, we assume the postman always starts from the root regardless of the room he had just visited.
#include<iostream>
#include<cstdio>
#include<cstring> using namespace std; const int maxn = 1e3+;
int a[maxn]; int main()
{
int c, n, q, x, num, d;
scanf("%d", &c);
while(c--)
{
memset(a, , sizeof(a)); scanf("%d", &n);
for(int i = ; i < n; i++)
scanf("%d", &a[i]);
scanf("%d", &q);
while(q--)
{
scanf("%d", &x);
if(x == a[])
{
puts("");
continue;
}
num = a[];
d = x - num; for(int i = ; i < n; i++)
{
if(d > )
{
while(a[i] < num) // 跳过所有与当前目标方向相反的点
i++;
printf("W");
num = a[i];
d = x - num;
}
else
{
while(a[i] > num)
i++;
printf("E");
num = a[i];
d = x - num;
}
if(a[i] == x)
break;
}
puts("");
}
}
return ;
}
写代码的时候就好好写代码……………………………………………………………………………………………………………………………………………………………………
Elven Postman的更多相关文章
- Elven Postman(BST )
Elven Postman Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- hdu 5444 Elven Postman 二叉树
Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Problem Descrip ...
- hdu 5444 Elven Postman
题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Description Elves are very peculia ...
- hdu 5444 Elven Postman(二叉树)——2015 ACM/ICPC Asia Regional Changchun Online
Problem Description Elves are very peculiar creatures. As we all know, they can live for a very long ...
- Hdu 5444 Elven Postman dfs
Elven Postman Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid= ...
- Elven Postman(二叉树)
Elven Postman Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- (二叉树)Elven Postman -- HDU -- 54444(2015 ACM/ICPC Asia Regional Changchun Online)
http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limit: 1500/1000 MS (Java/Others) ...
- HDU 5444 Elven Postman (2015 ACM/ICPC Asia Regional Changchun Online)
Elven Postman Elves are very peculiar creatures. As we all know, they can live for a very long time ...
- 2015 ACM/ICPC Asia Regional Changchun Online HDU 5444 Elven Postman【二叉排序树的建树和遍历查找】
Elven Postman Time Limit: 1500/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)T ...
- hdu 5444 Elven Postman(长春网路赛——平衡二叉树遍历)
题目链接:pid=5444http://">http://acm.hdu.edu.cn/showproblem.php?pid=5444 Elven Postman Time Limi ...
随机推荐
- Microsoft SQL Server 2008 R2官方中文版(SQL2008下载)
Microsoft SQL Server 2008 R2官方中文版(SQL2008下载) http://www.2cto.com/database/201308/235349.html
- Tclientdataset的CloneCursor问题 clientdataset 复制 赋值 的问题
http://www.myexception.cn/delphi/720245.html Tclientdataset的CloneCursor问题我的DBGRID所连接的数据源为cdsBook,然后用 ...
- 16/7/8_PHP-字符串介绍
不知道为什么慕课网还这个都要介绍.不过还是一个新的知识点. PHP开发中,我们遇到最多的可能就是字符串. 字符串变量用于包含字符串的值. 一个字符串 通过下面的3种方法来定义: 1.单引号2.双引号3 ...
- 工具 - MSF
#ms17- use auxiliary/scanner/smb/smb_ms17_010 - exploit use exploit/windows/smb/ms17_010_eternalblue ...
- [Python3 填坑] 018 组装类的几个例子
目录 1. print( 坑的信息 ) 2. 开始填坑 2.1 MetaClass 举例 2.2 type 举例 2.3 MetaClass 举例 1. print( 坑的信息 ) 挖坑时间:2019 ...
- 使用TreeSet对象去重排序
java对象除了采用equals和hashCode判断对象是否相等外,开发人员也可以通过其他属性判断两个对象是否相等 以下案例采用TreeSet去掉重复对象 Teacher: public class ...
- SQL取年月日的不同格式
Select CONVERT(varchar(100), GETDATE(), 0): 05 16 2006 10:57AM Select CONVERT(varchar(100), GETDATE( ...
- 到底如何设置 Java 线程池的大小?
来源:ifeve.com/how-to-calculate-threadpool-size/ 在我们日常业务开发过程中,或多或少都会用到并发的功能.那么在用到并发功能的过程中,就肯定会碰到下面这个问题 ...
- ARM汇编1
一. 指令和伪指令 1.1. 指令 a. (汇编)指令是CPU机器指令的助记符,经过编译后会得到一串10组成的机器码,可以由CPU读取执行. 1.2. 伪指令 b. (汇编)伪指令本质上不是指令(只是 ...
- FZUOJ-2275 Game
Problem 2275 Game Accept: 159 Submit: 539 Time Limit: 1000 mSec Memory Limit : 262144 KB Pro ...