519. 3D City Model
Time limit per test: 0.25 second(s)Memory limit: 262144 kilobytes
input: standardoutput: standard
grid where all the grid cells are equal squares. Each of thegrid cells can serve as a foundation of a single building in the city. A building is represented as a number of 1 x 1 x 1 cubes stacked on the top of each other. The cube that lays in the foundation of a building entirely occupies a single cell on the grid. It is clear that adjacent buildings can share a wall or a part of it. Typical cities can be seen on the image below.
The King of Berland has a 3D model of the capital city in his office. This model was made on a special 3D-printer out of plastic. It represents a layout of the capital city, but the scale is smaller, so it’s very convenient for the King to examine the model without having to visit the city itself. The King is bored though because the model is colorless, so he wants to paint the model. To calculate the exact amount of required paint he should know the total area of the model’s surface.You have to help the King and write a program that will calculate the required surface area of the given model. While calculating the surface area you should count not only the side surfaces, but also the areas of the top and bottom facets.matrix of digits. A digit in thej-th position of thei-th row stands for the height of the building with its foundation in cell () of the model. If the corresponding digit is equal to "0", it means there is no building built on the top of this cell.
Input
(1 ≤≤ 100), wheren— amount of rows in the given grid,m— amount of columns. The followingnlines contain the description of the model. Thesenlines containmdigits each representing heights of the buildings. It’s guaranteed that the given matrix contains at least one non-zero digit.
Output
Output the only positive integer — surface area of the model.
Example(s)
sample inputsample output
3 311121211138
sample inputsample output
3 410000010000012
Note
The first sample test corresponds to the leftmost picture from the problem statement.
AC代码：
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int n, m;int map[105][105];int xx[4] = {-1, 0, 1, 0}, yy[4] = {0, 1, 0, -1};int fun(int x, int y) {int ret = 0;if(map[x][y] == 0) return 0;if(map[x][y] > 0) ret += 2;for(int i = 0; i < 4; i++) {if(map[x][y] > map[x + xx[i]][y + yy[i]]) ret += (map[x][y] – map[x+xx[i]][y+yy[i]]);}return ret;}int main() {while(scanf("%d %d", &n, &m) != EOF) {memset(map, 0, sizeof(map));for(int i = 1; i <= n; i++) {char tmp[105];scanf("%s", tmp);int len = strlen(tmp);for(int j = 0; j < len; j++)map[i][j+1] = tmp[j] – '0';}int ans = 0;for(int i = 1; i <= n; i++) {for(int j = 1; j<= m; j++) {ans += fun(i, j);}}printf("%d\n", ans);}return 0;}
，生命太过短暂，今天放弃了明天不一定能得到
