What's the minimum number of people required? [closed]

Now there is a $n\times n$ square matrix, and then take $m$ rows and $m$ columns randomly. At least one intersection stood a man. How many people need to stand in this square matrix at least?

The original problem is an interesting puzzle, I'll express it in the form of a story.

A Clever King and a Condemned Man
Once upon a time,a famous condemned man is about to be executed.But the clever King didn't want him to die hastily. So before the execution, the king pointed to the formation in front of him and said:
'Now there is a formation which has 7 rows and 7 columns,You choose 33 soldiers and let them leave the formation,The remaining soldiers stand still, and then I choose 3 rows and 3 columns.If at least one soldier stands at the 9 intersections, you will be free,otherwise, the excution will continue.'
So how should the condemned man arrange the soliders to ensure that he will not be executed?

When I saw this puzzle, I guessed that the solution must follow some symmetry, so there must be 2 soldiers in each row and column, but I found that no matter how I arranged, at least 17 soldiers were required. Then I simulated a solution with computer (as shown in the figure below),then I found that the solution of only using 16 soliders exists and the solution was asymmetric, so my idea was wrong.

One of the solutions: $\begin{array}{1} 1&0&0&0&1&0&0\\ 1&1&0&0&0&0&1\\ 0&0&1&0&0&0&0\\ 1&0&0&1&0&1&0\\ 0&1&0&0&0&1&0\\ 0&0&0&1&0&0&1\\ 0&0&0&0&1&1&1 \end{array} $

According to the answer,I have been conscious that this question is equivalent to Zarankiewicz problem. Except for few cases, no satisfactory solution to the problem is known.And if we know the value of Z(n,n,m,m),we will get the answer to my problem:
$n^2-Z(n,n,m,m)+1$.

The solution to the story's specific puzzle ($n=7,m=3$) is in fact both optimal and unique up to permutations of rows and columns; here's its "nicest" presentation: $$\begin{bmatrix} &&1&1&&&\\ &1&&&1&&\\ 1&&&&&1&\\ 1&&&1&1&&\\ &1&&1&&1&\\ &&1&&1&1&\\ &&&&&&1 \end{bmatrix}$$ The general problem is extremely hard (Zarankiewicz's problem). For example, let $a_n$ be the solution for an $n×n$ board with $m=3$, then the $a_n$ follow no discernible pattern (and hence have been included in the OEIS as A350237): $$\{a_n\}_{n\ge1}=(0,0,1,3,5,10,16,22,32,40,52,64\dots)$$ Here are corresponding optimal solutions (all are also unique up to row/column permutations and transpose, except that there are $7$ non-isomorphic solutions for $n=9$):

O.. O... O.... O..... ..OO... ..OO.O.. .OO.OOO.. OO..OO.... .OOOO......
... .O.. .O... .OO... .O..O.. ...OO.O. OO..O..O. OOO...O... OO....O...O
... ..O. ..O.. .O.O.. O....O. O...OO.. O.O.O...O .OOO...O.. O...O.OOO..
    .... ...O. ..O.O. O..OO.. .O...OO. ...OO.O.. ..OOO...O. O..O.O..O..
         ....O ...OO. .O.O.O. O.O...O. OOOO..... O..OO....O O.O.O..O.O.
               .....O ..O.OO. OO.O.... O....O... O....O.OO. ...O..OO.OO
                      ......O .OO.O... O..O...OO .O....O.OO .OO..OOO...
                              .......O .O....O.O ..O..O.O.O ..O.OOO.O.O
                                       ..O...OO. ...O.OO.O. ..OO...OO.O
                                                 ....O.OO.O ....OO...OO
                                                            .O...O.OOO.

I've also obtained explicit upper bounds for $a_{13}$ to $a_{16}$, and hence lower bounds for Zarankiewicz at those points: $$a_{13}\le77,a_{14}\le91,a_{15}\le105,a_{16}\le128$$ The matrix giving the bound at $a_{16}$ in particular can be arranged like a snowflake (really four Walsh matrices inverted and placed together):

11111111........
1.1.1.1.1.1.1.1.
11..11..11..11..
1..11..1.11..11.
1111....1111....
1.1..1.1.1.11.1.
11....11..1111..
1..1.11.1..1.11.
.11.1..1.11.1..1
..1111..11....11
.1.11.1.1.1..1.1
....1111....1111
.11..11.1..11..1
..11..11..11..11
.1.1.1.1.1.1.1.1
........11111111