L – House Building

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hdu.edu.cn/showproblem.php?pid=5538

Description

 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   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   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   rows and   columns on the ground, an intersection of a row and a column is a   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    . Which     indicates the height of his house on the square of  -th row and  -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).
 

Under two situations the player could score one point.

⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won’t get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.

⋅2. Ignoring the buoys and relying on dogfighting to get point. If you and your opponent meet in the same position, you can try to fight with your opponent to score one point. For the proposal of game balance, two players are not allowed to fight before buoy #2 is touched by anybody.

There are three types of players.

Speeder: As a player specializing in high speed movement, he/she tries to avoid dogfighting while attempting to gain points by touching buoys.
Fighter: As a player specializing in dogfighting, he/she always tries to fight with the opponent to score points. Since a fighter is slower than a speeder, it’s difficult for him/her to score points by touching buoys when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.

There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.

Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting. Since Asuka is slower than Shion, she decides to fight with Shion for only one time during the match. It is also assumed that if Asuka and Shion touch the buoy in the same time, the point will be given to Asuka and Asuka could also fight with Shion at the buoy. We assume that in such scenario, the dogfighting must happen after the buoy is touched by Asuka or Shion.

The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?

Input

The first line contains an integer T 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

Output

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

Sample Input

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

Sample Output

30
20

HINT

题意

 给你一个n*m的空地，空地上有些地方有高为ai的建筑，然后让你在这些建筑的表面涂颜色，问你得花费多少？

底面不算

题解：

直接暴力，水题

代码

#include<iostream>
#include<stdio.h>
#include<cstring>
using namespace std;

int a[55][55];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int main()
{
    int t;scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        memset(a,0,sizeof(a));
        int n,m;scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                scanf("%d",&a[i][j]);
        int ans = 0;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                if(a[i][j])
                    ans++;
                for(int k=0;k<4;k++)
                {
                    int x = i+dx[k];
                    int y = j+dy[k];
                    if(a[i][j]>a[x][y])
                        ans+=a[i][j]-a[x][y];
                }
            }
        }
        printf("%d\n",ans);
    }
}