8 A $3 \times 3$ square with a unit square removed from each of its four corners is called a "cross shape".

On a $10 \times 11$ chessboard, what is the maximum number of non-overlapping "cross shapes" that can be placed? (Each "cross shape" exactly covers 5 small squares on the chessboard.)
Solution: 15.

First, prove that the maximum number of "crosses" that can be placed is 15. Use proof by contradiction, assuming that 16 "crosses" can be placed.

For each "cross," we call the central square the "heart" (denoted as *). Remove the outermost layer of squares from the $10 \times 11$ chessboard, leaving an $8 \times 9$ grid. Clearly, the "heart" of each "cross" can only appear in this $8 \times 9$ grid.
Note that in every $3 \times 3$ grid, at most two "hearts" can be placed:
$$
\begin{array}{l}
\Rightarrow \\
\Rightarrow
\end{array}
$$

We now discuss how to place as many "hearts" as possible in an $8 \times 3$ grid. Divide the $8 \times 3$ grid from top to bottom into three sub-grids: $3 \times 3$, $2 \times 3$, and $3 \times 3$. If one "heart" is placed in the middle $2 \times 3$ grid, then since at most two "hearts" can be placed in each $3 \times 3$ grid, a total of 5 "hearts" can be placed. If two "hearts" are placed in the middle $2 \times 3$ grid, then a total of 6 "hearts" can be placed in the $8 \times 3$ grid, and the third and sixth rows must be empty. Through exhaustive enumeration, it is known that there are only two ways to place 6 "hearts" as shown in the following diagram:

Now, divide the $8 \times 9$ grid into three $8 \times 3$ grids, from left to right, called grid (a), grid (b), and grid (c). Since there are 16 "hearts" in the $8 \times 9$ grid, one of the $8 \times 3$ grids must have 6 "hearts".

If grid (b) has 6 "hearts," since there are only two mutually symmetrical ways to place 6 "hearts" in an $8 \times 3$ grid, we can assume the 6 "hearts" are distributed as shown in the left diagram below:
It is easy to see that the 3rd and 7th columns cannot have "hearts." Thus, it is not difficult to see that grid (a) and grid (c) can each have at most 4 "hearts," resulting in a total of $4+6+4=14$ "hearts," leading to a contradiction.
If grid (a) and grid (c) each have 6 "hearts," then the 4th and 6th columns cannot have "hearts," so grid (b) can only have "hearts" in the middle column. It is not difficult to see that grid (b) can have at most 3 "hearts," resulting in a total of $6+3+6=15$ "hearts," also leading to a contradiction.
If one of grid (a) and grid (c) has 6 "hearts," assume grid (a) has 6 "hearts," and they are placed as shown in the right diagram above. In this case, the 4th column cannot have "hearts," so grid (b) can have at most 4 "hearts," resulting in a total of $6+4+5=15$ "hearts," also leading to a contradiction.
In summary, it is known that in the $8 \times 9$ grid, at most 15 "hearts" can be placed. Finally, in addition to the two examples already given, we provide two more examples to show that 15 "hearts" can indeed be placed:
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