HDU 5538 (水不水?)
House Building
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 248 Accepted Submission(s): 178
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.
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).
indicating the total number of test cases.
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
20

#include<bits/stdc++.h>
using namespace std;
int t;
int dis[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
int n,m;
int exm[55][55];
int fu;
int re;
int main()
{
while(scanf("%d",&t)!=EOF)
{
for(int i=1; i<=t; i++)
{
memset(exm,0,sizeof(exm));
fu=0;
re=0;
scanf("%d%d",&n,&m);
for(int j=1; j<=n; j++)
for(int k=1; k<=m; k++)
{
scanf("%d",&exm[j][k]);
if(exm[j][k]>=1)
fu++;
}
re+=fu;
for(int j=1;j<=n;j++)
for(int k=1;k<=m;k++)
{
for(int l=0;l<4;l++)
{
int xx=j+dis[l][0];
int yy=k+dis[l][1];
if(exm[j][k]>exm[xx][yy])
re=re+(exm[j][k]-exm[xx][yy]);
}
}
printf("%d\n",re);
}
}
return 0;
}
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