C++ 2 Dimensional Map 找到给定索引的所有相邻方 block

Question

我很难想出一种有效的方法来找到二维容器中给定值的所有相邻方 block 。假设我有一个容器表示为:

. . . . .
. G . . .
. . . . .
. . . . .
. . . . G

现在在我的程序中生成容器后(如上所示)，我需要将 G 的所有相邻方 block 设置为 X，因此 map 应该看起来喜欢:

X X X . .
X G X . .
X X X . .
. . . X X
. . . X G

我觉得有一种方法比我目前处理此解决方案的方法简单得多。这是我的思考过程:

for r to container.size()
    for c to container.size()
        if r == 0                            //if we're at the first row, don't check above
            if c == 0                        //if we're at the first column, don't check to the left

            else if c == container.size() -1 //if we're at the last column, don't check to the right

        else if r == container.size()        //if we're at the last row, don't check below
            if c == 0                        //if we're at the first column, don't check to the left

            else if c == container.size() -1 //if we're at the last column, don't check to the right

        else if c == 0                       //if we're at the first column, don't check to the left

        else if c == container.size() - 1    //if we're at the last column, don't check to the right

        else                                 //check everything

这似乎是 realllllllly 重复，但是，我必须检查很多条件以避免意外检查到我的 map 边界之外。我将为每个单独的 if/编写 lot 的 if(map[r][c+1] == "G") 语句else 写在上面。 有没有其他方法可以检查正方形是否与我没有想到的 G 相邻？

Answer 1

简化此操作的一种方法是为您的数据结构添加边框，因此您有例如

char map[N+2][N+2];

然后像这样遍历 map :

for (i = 1; i <= N; ++i)
{
    for (j = 1; j <= N; ++j)
    {
        if (map[i][j] == 'G')
        {
            for (di = -1; di <= 1; ++di)
            {
                for (dj = -1; dj <= 1; ++dj)
                {
                    if (di == 0 && dj == 0)      // skip central value
                        continue;
                    map[i + di][j + dj] = 'X';   // set neighbouring values to 'X'
                }
            }
        }
    }
}