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HDU_5538_House Building

2018-06-17 21:48:23来源：未知阅读 ()

House Building

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1958 Accepted Submission(s): 1223

Problem 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

Figure 1: A typical world in Minecraft.

Input

The first line contains an integer

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

Figure 2: A top view and side view image for sample test case 1.

Source

2015ACM/ICPC亚洲区长春站-重现赛（感谢东北师大）

计算立方体表面积
对于给出的每个高度，先算排除底面的表面积，然后减去旁边四面高度的min就是当前立方体表面积

 1 #include <iostream>
 2 #include <string>
 3 #include <cstdio>
 4 #include <cstring>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <cmath>
 8 #include <vector>
 9 #include <queue>
10 #include <stack>
11 #include <set>
12 #include <map>
13 using namespace std;
14 typedef long long           LL ;
15 typedef unsigned long long ULL ;
16 const int    maxn = 1e3 + 10   ;
17 const int    inf  = 0x3f3f3f3f ;
18 const int    npos = -1         ;
19 const int    mod  = 1e9 + 7    ;
20 const int    mxx  = 100 + 5    ;
21 const double eps  = 1e-6       ;
22 const double PI   = acos(-1.0) ;
23 
24 int T, n, m, a[maxn][maxn];
25 LL ans;
26 int main(){
27     // freopen("in.txt","r",stdin);
28     // freopen("out.txt","w",stdout);
29     while(~scanf("%d",&T)){
30         while(T--){
31             ans=0;
32             memset(a,0,sizeof(a));
33             scanf("%d %d",&n,&m);
34             for(int i=1;i<=n;i++)
35                 for(int j=1;j<=m;j++)
36                     scanf("%d",&a[i][j]);
37             for(int i=1;i<=n;i++)
38                 for(int j=1;j<=m;j++)
39                     if(a[i][j])
40                         ans+=(4*a[i][j]+1)
41                             -min(a[i][j],a[i-1][j])-min(a[i][j],a[i+1][j])
42                             -min(a[i][j],a[i][j-1])-min(a[i][j],a[i][j+1]);
43             printf("%lld\n",ans);
44         }
45     }
46     return 0;
47 }