House Building（几何体表面积）

题目：

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 \le T \le 501≤T≤50

• 1\le n,m\le 501≤n,m≤50

• 0 \le c_{i,j} \le 10000≤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

分析：

首先计算出有多少个立方体，乘以6以后减去被遮盖的面的面积；有三种

1.底面的不计算；

2.上下的遮盖；

3.前后左右的覆盖；

1 2都很好解决，3用两次二维循环，判断连续的两个位置是否都不是零，如果都不是，那么被覆盖的面一定是盖度较低的*2；

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
int mapp[52][52];
int n,m,ans,cnt,dnf;
using namespace std;
int main()
{
    int t,i,j,k,a;
    scanf("%d",&t);
    while(t--)
    {
        ans=0;
        cnt=0;
        dnf=0;
        memset(mapp,0,sizeof(0));
        scanf("%d %d",&n,&m);
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
            {
                scanf("%d",&a);
                mapp[i][j]=a;
                ans+=a;//计算总立方体个数
                if(a!=0)
                    cnt++;//计算底面的面积
                if(a>1)
                    dnf=dnf+a-1;//计算上下面的遮盖数（*2）
            }
        ans=ans*6-cnt;
        for(i=1;i<=m;i++)
            for(j=1;j<n;j++)
                if((mapp[j][i]!=0)&&(mapp[j+1][i]!=0))
                        dnf+=min(mapp[j][i],mapp[j+1][i]);
        for(i=1;i<=n;i++)
            for(j=1;j<m;j++)
                if((mapp[i][j]!=0)&&(mapp[i][j+1]!=0))
                   dnf+=min(mapp[i][j],mapp[i][j+1]);
        ans=ans-dnf*2;
        printf("%d\n",ans);
    }
    return 0;
}