There is a $8 \times 8$ piece of land, with two pairs of opposite sides running east-west and north-south, respectively. Divide this land into 64 $1 \times 1$ squares. Build some rooms on this land so that each square has at most one room, and each room occupies only one square.

If a room has rooms on the east, west, and south sides of its square, then it is called a shaded room. How many rooms can be built on this land at most, so that none of the rooms are shaded?
Note: By definition, rooms located on the east, west, and south boundaries are not shaded.
Consider the land as an $8 \times 8$ chessboard. For each cell on the chessboard, if it corresponds to a square with a room, color it black; otherwise, color it white. We use $(i, j)$ to denote the cell in the $i$-th row and $j$-th column (consider the northernmost row as the first row and the westernmost column as the first column). We first prove that there exists a coloring scheme that maximizes the number of black cells (called the optimal scheme) in which all cells on the eastern, western, and southern borders are colored black.

Assume there exists an optimal scheme where the cell $(i, 1)$ is colored white. Since this is an optimal scheme, this cell cannot be colored black, meaning that if $(i, 1)$ were colored black, then $(i, 2)$ would become shaded. In other words, $(i, 2)$, $(i, 3)$, and $(i+1, 2)$ are all colored black. At this point, we can swap the colors of $(i, 1)$ and $(i, 2)$, and the number of black cells remains unchanged, while $(i, 1)$ is now colored black, with no impact on other cells.

Similarly, for all cells on the eastern, western, and southern borders (i.e., from $(1,1)$ to $(8,1)$, from $(8,1)$ to $(8,8)$, and from $(8,8)$ to $(1,8)$), we can make the above adjustments. This shows that there exists an optimal scheme in which all cells on the eastern, western, and southern borders are colored black.

Next, consider the remaining $7 \times 6$ sub-chessboard. We prove that the number of black cells in this sub-chessboard does not exceed 28. To achieve 28 black cells, the average number of black cells per row must be 4. In fact, if a row has 6 black cells, then the next row cannot have any black cells (such black cells would make the cells directly north of them shaded). If a row has 5 black cells, then the next row can have at most 3 black cells (the cell directly below the only white cell and the two adjacent cells). This indicates that the average number of black cells per row in the $7 \times 6$ sub-chessboard does not exceed 4, so it contains at most 28 black cells.

Thus, the upper bound for the number of black cells is 28, which includes the 22 cells on the eastern, western, and southern borders and the 28 cells in the remaining $7 \times 6$ sub-chessboard, totaling 50 cells.

Below is an example where the number of black cells is 50. Color the 1st, 2nd, 4th, 5th, 7th, and 8th columns black, and color the 8th row black, with the rest of the cells colored white. It is easy to verify that this coloring scheme meets the requirements and has exactly 50 black cells.

In conclusion, the land can accommodate at most 50 rooms such that none of the rooms are shaded.