Have you ever played the video game Minecraft? This game has been one of the world's most popular game in recent years. The world of Minecraft is made up of lots of 1×1×1 blocks in a 3D map. Blocks are the basic units of structure in Minecraft, there are many types of blocks. A block can either be a clay, dirt, water, wood, air, ... or even a building material such as brick or concrete in this game.

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\times mn×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 \le i\le n,1\le j\le m1≤i≤n,1≤j≤m). Which ci,j indicates the height of his house on the square of ii-th row and jj-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).

Input Format

The first line contains an integer TT indicating the total number of test cases.

First line of each test case is a line with two integers n,mn,m.

The n lines that follow describe the array of Nyanko-san's blueprint, the ii-th of these lines has mm integers ci,1,ci,2,...,ci,m, separated by a single space.

1≤T≤50

1≤n,m≤50

1≤ci,j​≤1000

Output Format

For each test case, please output the number of glass units you need to collect to meet Nyanko-san's requirement in one line.

HINT

样例输入复制

2
3 3
1 0 0
3 1 2
1 1 0
3 3
1 0 1
0 0 0
1 0 1

样例输出复制

30
20

题目来源

ACM Changchun 2015

//空间立方体(由多个1*1*1的小立方体组成)的表面积

 int t,n,m;
int a[][];
int dir[][]={,,,-,,,-,};
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
int ans=;
memset(a,,sizeof(a));//一定要清0
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
scanf("%d",&a[i][j]);
if(a[i][j]) ans++;//底面
}
}
for(int i=;i<=n;i++)
{
for(int j=;j<=m;j++)
{
for(int k=;k<;k++)
{
int x=i+dir[k][];
int y=j+dir[k][];
if(a[i][j]>a[x][y]) ans+=a[i][j]-a[x][y];
}
//(i,j)的block 在4个方向上对答案的贡献
}
}
printf("%d\n",ans);
}
return ;
}

ACM Changchun 2015 L . House Building的更多相关文章

  1. ACM Changchun 2015 J. Chip Factory

    John is a manager of a CPU chip factory, the factory produces lots of chips everyday. To manage larg ...

  2. ACM Changchun 2015 A. Too Rich

    You are a rich person, and you think your wallet is too heavy and full now. So you want to give me s ...

  3. 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 ...

  4. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 G. Garden Gathering

    Problem G. Garden Gathering Input file: standard input Output file: standard output Time limit: 3 se ...

  5. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 D. Delay Time

    Problem D. Delay Time Input file: standard input Output file: standard output Time limit: 1 second M ...

  6. 【第13届景驰-埃森哲杯广东工业大学ACM程序设计大赛-L】用来作弊的药水

    链接:https://www.nowcoder.com/acm/contest/90/L来源:牛客网 输入x,a,y,b,(1<=x,a,y,b<=10^9)判断x^a是否等于y^b 前面 ...

  7. 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 ...

  8. ACM ICPC 2015 Moscow Subregional Russia, Moscow, Dolgoprudny, October, 18, 2015 I. Illegal or Not?

    I. Illegal or Not? time limit per test 1 second memory limit per test 512 megabytes input standard i ...

  9. 2019山东ACM省赛L题题解(FLOYD传递闭包的变形)

    本题地址 https://cn.vjudge.net/contest/302014#problem/L Median Time Limit: 1 Second      Memory Limit: 6 ...

随机推荐

  1. F. Coprime Subsequences 莫比乌斯反演

    http://codeforces.com/contest/803/problem/F 这题正面做了一发dp dp[j]表示产生gcd = j的时候的方案总数. 然后稳稳地超时. 考虑容斥. 总答案数 ...

  2. vi命令使用

    在vi下如何显示行号? 按Esc切换到命令行模式,输入: :set nu 如果您想每次进入vi都标出行号,编辑~/.vimrc文件.也就是在用户的主目录下,编辑存档.vimrc文件.里边写一行: se ...

  3. [摘录]全面学习GFW

    转载自:https://cokebar.info/archives/253 GFW会是一个长期的存在.要学会与之共存,必须先了解GFW是什么.做为局外人,学习GFW有六个角度.渐进的来看分别是: 首先 ...

  4. C#里边的控件缩写大全(比较规范)

    标准控件1 btn Button 2 chk CheckBox 3 ckl CheckedListBox 4 cmb ComboBox 5 dtp DateTimePicker 6 lbl Label ...

  5. NOPI Excel 读取公式生成后的数据

    using NPOI.HSSF.UserModel; using NPOI.SS.UserModel; using NPOI.XSSF.UserModel; using System; using S ...

  6. Memcache笔记02-telnet操作memcached

    telnet操作Memcached 登录到telnet连接到memcached服务: telnet 127.0.0.1 11211 memcached的基本命令: //当telnet登录成功可以看到一 ...

  7. ABAP和Java单例模式的攻防

    ABAP CLASS zcl_jerry_singleton DEFINITION PUBLIC FINAL CREATE PRIVATE . PUBLIC SECTION. INTERFACES i ...

  8. 第011课_串口(UART)的使用

    from: 第011课_串口(UART)的使用 第001节_辅线1_硬件知识_UART硬件介绍 1. 串口的硬件介绍 UART的全称是 Universal Asynchronous Receiver ...

  9. HTTP、HTTP2.0、HTTPS、SPDY

    本文原链接:https://cloud.tencent.com/developer/article/1082516 HTTP,HTTP2.0,SPDY,HTTPS你应该知道的一些事 1.web始祖HT ...

  10. 阻止form元素内的input标签回车提交表单

    <form></form>标签内input元素回车会默认提交表单. 阻止回车默认提交表单: $('form').on('keydown', function (event) { ...