Manhattan 2025
Time Limit: 1000MS   Memory Limit: 30000K
Total Submissions: 1318   Accepted: 703

Description

Background Manhattan in the year 2025 - it is so densely populated that its old two-dimensional grid of streets and avenues fails to provide enough space for all the traditional vehicles such as cars, bicycles, or busses.Accordingly, the newly developed 3D-Skyjetters become very popular, because they allow to pass the traffic jams on the ground by flying in the air. After a series of horrible accidents caused by 3D-Skyjetters cutting a corner, New York authorities have put into place new regulations of air traffic and are determined to enforce them rigorously. The key point of these regulations is that 3D-Skyjetters must follow virtual airways on a three-dimensional rectangular grid, easy enough for the New Yorkers who had to use the two-dimensional rectangular grid of roads on the ground all their life. Problem You own a company that rents out 3D-Skyjetters. As 3D-Skyjetters are in such an early state of development,they are far from being economical. So your customers keep running out of petrol at all the wrong places,and you need a system to inform them about their current range at all times. You may assume that travelling from one intersection to the next in the grid takes one unit of petrol, no matter if the customer is moving horizontally or vertically, up or down. You may also assume that your customer is located at some intersection where his or her movements are not restricted by the ground or other obstacles, but just by the amount of remaining petrol. Given the amount of petrol, provide a graphical representation of all the intersections in the range of your customer, along with the amount of petrol that is needed to go there.

Input

The first line of the input contains the number of scenarios. For each scenario, there is a line containing the units of remaining petrol, i.e an integer u satisfying 0 <= u <= 9. If more than 9 units of petrol remain, the customer will ignore the display anyway.

Output

Start the output for each scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a graphical (or rather textual) representation of all intersections that can be reached with the given amount of petrol, along with the units of petrol necessary to go there. In this graphical representation, print the slices of the smallest axis-aligned three-dimensional cube containing all the intersections in the range, and label the slices from the bottom to the top starting at 1. For each slice,start the output with a line containing "slice #s:", where s is the number of the slice. In the lines that follow, print a graphical representation of all the intersections in that slice, using

  • the digits 0 to 9 for intersections in the range, representing the amount of petrol necessary to go there,
  • and the dot "." for intersections not in the range.

Print an additional blank line after each scenario.

Sample Input

2
0
2

Sample Output

Scenario #1:
slice #1:
0 Scenario #2:
slice #1:
.....
.....
..2..
.....
.....
slice #2:
.....
..2..
.212.
..2..
.....
slice #3:
..2..
.212.
21012
.212.
..2..
slice #4:
.....
..2..
.212...2.. ..... slice #5: ..... ..... ..2.. ..... .....

Source

TUD Programming Contest 2003, Darmstadt, Germany
#include <stdio.h>
#include <string.h>
#include <stdlib.h> int Scenario,Oil,Slice,M[][];
void Display()
{
int i,j;
for (i=;i<Slice;i++)
{
for (j=;j<Slice;j++)
{
if (M[i][j]<=Oil)
printf("%d",M[i][j]);
else
printf(".");
}
printf("\n");
}
}
void left_up(int x,int y,int m,int t)
{
if (m==Oil+)
return;
M[x][y]=m;
left_up(x-,y,m+t,t);
left_up(x,y-,m+t,t); }
void up_right(int x,int y,int m,int t)
{
if (m==Oil+)
return;
M[x][y]=m;
up_right(x+,y,m+t,t);
up_right(x,y-,m+t,t);
}
void right_down(int x,int y,int m,int t)
{
if (m==Oil+)
return;
M[x][y]=m;
right_down(x+,y,m+t,t);
right_down(x,y+,m+t,t); }
void down_left(int x,int y,int m,int t)
{
if (m==Oil+)
return;
M[x][y]=m;
down_left(x,y+,m+t,t);
down_left(x-,y,m+t,t);
}
void func(int x,int y,int m,int t)
{
left_up(x,y,m,t);
up_right(x,y,m,t);
right_down(x,y,m,t);
down_left(x,y,m,t);
}
int main()
{
int i,j,x,y;
scanf("%d",&Scenario);
for(i=;i<=Scenario;i++)
{
scanf("%d",&Oil);
printf("Scenario #%d:\n",i);
if (Oil==)
{
printf("slice #1:\n0\n");
}
else
{
Slice=*Oil+;
for (j=;j<Slice;j++)
{
memset(M,,sizeof(M));
x=Oil;
y=Oil;
printf("slice #%d:\n",j+);
if (j<=Oil)
func(x,y,Oil-j,);
else
func(x,y,j-Oil,);
Display();
}
}
printf("\n");
}
return ;
}

注意方位的把握~~~

poj 1806 分块模拟的更多相关文章

  1. P1972 [SDOI2009]HH的项链[离线+树状数组/主席树/分块/模拟]

    题目背景 无 题目描述 HH 有一串由各种漂亮的贝壳组成的项链.HH 相信不同的贝壳会带来好运,所以每次散步完后,他都会随意取出一段贝壳,思考它们所表达的含义.HH 不断地收集新的贝壳,因此,他的项链 ...

  2. poj 3077Rounders(模拟)

    转载请注明出处:viewmode=contents">http://blog.csdn.net/u012860063? viewmode=contents 题目链接:http://po ...

  3. POJ 1068 Parencodings 模拟 难度:0

    http://poj.org/problem?id=1068 #include<cstdio> #include <cstring> using namespace std; ...

  4. POJ 1036 Rails 模拟堆栈

    水题,主要是思路清晰,判断明确. 记x为A站最前方的车,y表示下一列要进入B站的车厢,初识时,x=1;y=a1;C=[]; 在调度过程中: if(y==0)那么调度成功,退出模拟过程:否则 if(x= ...

  5. POJ 1001 Exponentiation 模拟小数幂

    模拟小数幂 小数点位 pos 非零末位 e 长度 len 只有三种情况 pos > len pos < e e < pos < len #include <iostrea ...

  6. A Simple Problem with Integers POJ - 3468 (分块)

    题目链接:https://cn.vjudge.net/problem/POJ-3468 题目大意:区间加减+区间查询操作. 具体思路:本来是一个线段树裸题,为了学习分块就按照分块的方法做吧. 分块真的 ...

  7. POJ 1008 简单模拟题

    e.... 虽然这是一道灰常简单的模拟题.但是米做的时候没有读懂第二个日历的计时方法.然后捏.敲完之后华丽的WA了进一个点.坑点就在一年的最后一天你是该输出本年的.e ...但是我好想并没有..看di ...

  8. poj 1806 Frequent values(RMQ 统计次数) 详细讲解

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1806 题目大意:给你一个非降序排列的整数数组,你的任务是对于一系列的询问,(i,j),回答序列中出现次 ...

  9. Crashing Robots POJ 2632 简单模拟

    Description In a modernized warehouse, robots are used to fetch the goods. Careful planning is neede ...

随机推荐

  1. kendoui treeview grid spreadsheet

    treeview 傻子方式获取id <!DOCTYPE html> <html> <head> <title>API</title> < ...

  2. MyElipes遇到 source not found解决方案(查看.class文件源码一劳永逸的解决方法)

    在用Myeclipse 或者是eclipse进行开发时候经常遇到这个问题. File class editor source not found 问题.原因很简单,就是因为这是一个源码包,相应的没有编 ...

  3. My安卓知识2--使用listview绑定sqlite中的数据

    我想在我的安卓项目中实现一个这样的功能,读取sqlite数据库中的数据并显示到某个页面的listview控件中. 首先,我建立了一个Service类,来实现对数据库的各种操作,然后在这个类中添加对数据 ...

  4. 有效提高win7电脑网速50倍

    有效提高win7电脑网速50倍 目前有多少人在使用Windows7?据统计微软已成功售出3.5亿份正版操作系统,加上市场上流传以Win7旗舰版为首的盗版Win7系统下载,Win7激活次数可能已经超过1 ...

  5. Getting Started With Hazelcast 读书笔记(第一章)

    第一章:数据集群的演化与 早期的服务器架构 显然,应用是可扩展的,但是由于是集中式服务器,随着数据库性能达到极限,再想扩展就变得极端困难,于是出现了缓存.    缓存显然再次提升了可扩展性,减轻了数据 ...

  6. 为什么不能访问django自带的索引页

    通过HTTP://192.168.160.128:8000访问虚拟机上的django索引页出现“ 无法访问此网站 192.168.160.128 拒绝了我们的连接请求. ” 是什么原因呢?费了好大一番 ...

  7. 2016/10/28 很久没更了 leetcode解题 3sum问题进阶版4sum

    18. 4Sum Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c  ...

  8. R语言读取excel文件的3种方法

    R读取excel文件中数据的方法: 电脑有一个excel文件,原始的文件路径是:E:\R workshop\mydata\biom excel数据为5乘2阶矩阵,元素为                ...

  9. .Net 对App.config和Web.config的访问操作(增、删、读、改)

    一.首先引用Configuration 1)App.config如下: using System.Configuration;//若果还没有Configuration,右键引用文件夹添加引用,在.NE ...

  10. JIT动态编译器的原理与实现之Interpreter(解释器)的实现(三)

    接下来,就是要实现一个虚拟机了.记得编码高质量的代码中有一条:不要过早地优化你的代码.所以,也本着循序渐进的原则,我将从实现一个解释器开始,逐步过渡到JIT动态编译器,这样的演化可以使原理看起来更清晰 ...