c++ - 无数个九宫格组成的矩阵，随机选一个点为中心，计算出它周围的8个格子内最大且比它大的格子做下一个中心，有什么简洁优美的计算方法吗？

Answer 1

用一个二维数组保存每个格子上的数 这样格子a[i][j]上面的数（如果上面有数，不是边界）就是a[i-1][j]， 左上角的数是a[i-1][j-1] 右上角的数是a[i-1][j+1]，下一步怎么走的话就是拿这周围的几个数中最大的那个跟自己比较， 直到没有比他小的，用数组记录路径。

Answer 2

If you are searching grid by grid, it is easiest to use a large two-dimensional array, and just write the eight surrounding grids by hand

[x-1,y-1]
[x,y-1]
[x+1,y-1]
...
[x+1,y+1]

Very elegant.

If the data set is large and regular, you can consider optimization methods to skip some grids to speed up the search. Of course, that’s another topic

Answer 3

The vectors in eight directions can be listed in advance, thus avoiding writing eight repeated sections of code.

As shown below:

const int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
const int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
// ...
for (int i = 0; i < 8; ++i) {
    int cur_x = x + dx[i];
    int cur_y = y + dy[i];
    // ...
}