5. Given an integer $n \geqslant 2$. From an $n \times n$ grid

remove $n$ cells located in different rows and different columns, the resulting shape is called an $n \times n$ "sieve". $1 \times k (k \in$ $\mathbf{Z}_{+}$) or $k \times 1$ sub-grids are both called a "bar". For any $n \times n$ sieve $A$, partition it into several bars (i.e., these bars do not overlap and exactly cover the sieve $A$), in all possible partitions of $A$, the minimum number of bars is denoted as $m(A)$. When $A$ ranges over all $n \times n$ sieves, find the range of values for $m(A)$.
5. $m(A)=2 n-2$.

First, prove: $m(A) \leqslant 2 n-2$.
Consider all possible maximal $1 \times k$ bars. For a row, if the removed cell is not in the first or last column, then two bars are needed for that row. Otherwise, one bar is needed. Therefore, a total of $2 n-2$ bars are needed. Thus, $m(A) \leqslant 2 n-2$.
Next, prove: $m(A) \geqslant 2 n-2$.
Use induction on $n$.
For $n=2$, it is clear that at least two bars are needed.
Assume for any $(n-1) \times (n-1)$ board $A'$, $m(A') \geqslant 2(n-1)-2$.
Consider an $n \times n$ board $A$ where $n \geqslant 3$. Define a "cross" as the set of cells in a row and a column, with the intersection being a removed cell.
Each bar is in at least one cross.
Since each bar can contain at most $n-1$ cells, and there are $n(n-1)$ cells in total, at least $n$ bars are needed.

If exactly $n$ bars are used, then each bar is exactly $n-1$ cells long. Given that each row and each column has one removed cell, these $n$ bars must all be horizontal or all be vertical, which is impossible. Therefore, at least $n+1$ bars are needed.

By the pigeonhole principle, there is a cross that contains at least two bars. Remove this cross and merge the separated parts of the bars back into single bars. Now, there are at most $m(A)-2$ bars. This leaves an $(n-1) \times (n-1)$ board.
By the induction hypothesis,
$m(A)-2 \geqslant 2(n-1)-2 \Rightarrow m(A) \geqslant 2 n-2$.
In conclusion, $m(A)=2 n-2$.