HDU 5538 House Building(模拟——思维)
题目链接:
http://acm.hdu.edu.cn/showproblem.php?pid=5538

Figure 1: A typical world in Minecraft.
Nyanko-san is one of the diehard fans of the game, what he loves most is to build monumental houses in the world of the game. One day, he found a flat ground in some place. Yes, a super flat ground without any roughness, it's really a lovely place to build houses on it. Nyanko-san decided to build on a n×m big flat ground, so he drew a blueprint of his house, and found some building materials to build.
While everything seems goes smoothly, something wrong happened. Nyanko-san found out he had forgotten to prepare glass elements, which is a important element to decorate his house. Now Nyanko-san gives you his blueprint of house and asking for your help. Your job is quite easy, collecting a sufficient number of the glass unit for building his house. But first, you have to calculate how many units of glass should be collected.
There are n rows and m columns on the ground, an intersection of a row and a column is a 1×1 square,and a square is a valid place for players to put blocks on. And to simplify this problem, Nynako-san's blueprint can be represented as an integer array ci,j(1≤i≤n,1≤j≤m). Which ci,j indicates the height of his house on the square of i-th row and j-th column. The number of glass unit that you need to collect is equal to the surface area of Nyanko-san's house(exclude the face adjacent to the ground).
First line of each test case is a line with two integers n,m.
The n lines that follow describe the array of Nyanko-san's blueprint, the i-th of these lines has m integers ci,1,ci,2,...,ci,m, separated by a single space.
1≤T≤50
1≤n,m≤50
0≤ci,j≤1000
Figure 2: A top view and side view image for sample test case 1.题意描述:
输入矩阵的大小及矩阵,计算该3维立体结构的表面积(底面积除外)
解题思路:
模拟处理,细心即可。
AC代码:
#include<stdio.h>
#include<string.h>
int map[][];
int main()
{
int T,i,j,k,sum,n,m,tx,ty;
int next[][]={,,,,,-,-,};
scanf("%d",&T);
while(T--)
{
scanf("%d%d",&n,&m);
memset(map,,sizeof(map));
for(i=;i<=n;i++)
for(j=;j<=m;j++)
scanf("%d",&map[i][j]);
sum=;
for(i=;i<=n;i++)
{
for(j=;j<=m;j++)
{
if(map[i][j])
{
sum++;
for(k=;k<=;k++)
{
tx=i+next[k][];
ty=j+next[k][];
if(map[i][j] > map[tx][ty])
sum += (map[i][j] - map[tx][ty]);
}
}
}
}
printf("%d\n",sum);
}
return ;
}
HDU 5538 House Building(模拟——思维)的更多相关文章
- hdu 5538 House Building(长春现场赛——水题)
题目链接:acm.hdu.edu.cn/showproblem.php?pid=5538 House Building Time Limit: 2000/1000 MS (Java/Others) ...
- 2015ACM/ICPC亚洲区长春站 L hdu 5538 House Building
House Building Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others) ...
- 模拟+思维 HDOJ 5319 Painter
题目传送门 /* 题意:刷墙,斜45度刷红色或蓝色,相交的成绿色,每次刷的是连续的一段,知道最终结果,问最少刷几次 模拟+思维:模拟能做,网上有更巧妙地做法,只要前一个不是一样的必然要刷一次,保证是最 ...
- HDU 6187 Destroy Walls (思维,最大生成树)
HDU 6187 Destroy Walls (思维,最大生成树) Destroy Walls *Time Limit: 8000/4000 MS (Java/Others) Memory Limit ...
- HDU 5538 L - House Building 水题
L - House Building Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.ph ...
- HDU 5122 K.Bro Sorting(模拟——思维题详解)
题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5122 Problem Description Matt's friend K.Bro is an A ...
- HDU 5538/ 2015长春区域 L.House Building 水题
题意:求给出图的表面积,不包括底面 #include<bits/stdc++.h> using namespace std ; typedef long long ll; #define ...
- HDU 5510---Bazinga(指针模拟)
题目链接 http://acm.hdu.edu.cn/search.php?action=listproblem Problem Description Ladies and gentlemen, p ...
- HDU 5047 Sawtooth(大数模拟)上海赛区网赛1006
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5047 解题报告:问一个“M”型可以把一个矩形的平面最多分割成多少块. 输入是有n个“M",现 ...
随机推荐
- SQL奇技淫巧
1.SQL行列转换 问题:假设有张学生成绩表(tb)如下:姓名 课程 分数张三 语文 74张三 数学 83张三 物理 93李四 语文 74李四 数学 84李四 物理 94想变成(得到如下结果): 姓名 ...
- Effective Java 第三版——14.考虑实现Comparable接口
Tips <Effective Java, Third Edition>一书英文版已经出版,这本书的第二版想必很多人都读过,号称Java四大名著之一,不过第二版2009年出版,到现在已经将 ...
- Git详解之八:Git与其他系统
Git 与其他系统 世界不是完美的.大多数时候,将所有接触到的项目全部转向 Git 是不可能的.有时我们不得不为某个项目使用其他的版本控制系统(VCS, Version Control System ...
- Git详解之一:Git起步
起步 本章介绍开始使用 Git 前的相关知识.我们会先了解一些版本控制工具的历史背景,然后试着让 Git 在你的系统上跑起来,直到最后配置好,可以正常开始开发工作.读完本章,你就会明白为什么 Git ...
- 在vim中使用zencoding/Emmet
zencoding在vim上的插件已经改名为Emmet.vim 1. 安装,推荐使用vundle插件管理器安装,在~/.vimrc中,添加:Bundle 'Emmet.vim',输入命令vim +Bu ...
- 【官方文档】Nginx模块Nginx-Rtmp-Module学习笔记(三)流式播放Live HLS视频
源码地址:https://github.com/Tinywan/PHP_Experience HTTP Live Streaming(HLS)是由Apple Inc.实施的非常强大的流视频协议.HLS ...
- vscode 开发工具
做开发两年了,而我记忆力不太好,所以写代码得靠强大的编辑器提示. 陆陆续续使用了如 notepad++.dreamweaver.sublime text.webstorm.phpstorm.Atom等 ...
- [转载]MySQL UUID() 函数
目录 目录 一 引子 二 MySQL UUID() 函数 三 复制中的 UUID()四 UUID_SHORT() 函数 3.1 实验环境介绍 3.2 搭建复制环境 3.3 基于 STATEMENT 模 ...
- 利用jquery encoder解决XSS脚本注入所产生的问题
问题现象:前端接收到后台一个数据(其中包含html)标签,自动转译成html页面元素,且自动执行了脚本,造成了前端页面的阻塞 接受的后台数据为大量重复的如下代码 ");</script ...
- python入门之函数
为什么要用函数 python的函数是由一个新的语句编写,即def ,def是可执行的语句--函数并不存在,知道python运行了def后才存在. 函数是通过赋值函数传递的,参数通过赋值传递给函数. d ...