$$$a1$$$ $$$a2$$$ $$$a3$$$ $$$x$$$ $$$a5$$$ $$$a6$$$ $$$a7$$$
$$$b1$$$ $$$b2$$$ $$$b3$$$ $$$y$$$ $$$b5$$$ $$$b6$$$ $$$b7$$$
Lets assume numbers in position $$$i = 4$$$` $$$x$$$ and $$$y$$$ are swaped. $$$y$$$ may go to positions $$${4, 5, 6, 7}$$$. $$$y$$$ can't go to positions $$${1, 2, 3}$$$ cause $$$y > x >= a3 >= a2 >= a1$$$. Similarly $$$x$$$ may go to positions $$${1, 2, 3, 4}$$$ but can't go to positions $$${5, 6, 7}$$$. Its easy to see that still $$$a_1 <= b_1, a_2 <= b_2, a_3 <= b_3$$$, furthermore $$$a_4 <= b_4$$$ $$$=>$$$ Next swap will occurr in $$$j$$$, where $$$j > i$$$.

Now we need a data structure with fast $$$insert$$$, $$$erase$$$ and $$$get minimum$$$ operations. In C++ we have $$$std::multiset<int>$$$. We'll maintain two of them` $$$A$$$ and $$$B$$$. Lets iterate over positions $$$1, 2, ..., n$$$ in this order. When we get to position $$$i$$$, $$$A$$$ must contain $$$a_i, a_{i+1}, ..., a_n$$$, $$$B$$$ must contain $$$b_i, b_{i+1}, ..., b_n$$$. Note: $$$A.min$$$ is equal to $$$b_i$$$, $$$B.min$$$ is equal to $$$a_i$$$.
If $$$A.min > B.min$$$ we must swap $$$a_i$$$ and $$$b_i$$$. Lets erase $$$B.min$$$ from $$$B$$$, insert it into $$$A$$$. Erase $$$A.min$$$ and insert it into $$$B$$$. We know that next swap will occurr at position $$$j$$$, $$$j > i$$$ so we are done with position $$$i$$$. If $$$A.min <= B.min$$$ we are finished, too. So we just erase $$$A.min$$$ from $$$A$$$. $$$A.min$$$ is $$$a_i$$$ thus we are left with numbers $$$a_{i+1}, a_{i+2}, ..., a_n$$$. Do the same with $$$B$$$.
Now we must just count swaps and print it.

Reversing row $$$i$$$ $$$<=>$$$ reversing row $$$n - i + 1$$$
$$$i <= n / 2$$$ or $$$n - i + 1 <= n / 2$$$
Thus we can assume that we will only reverse rows $$$1, 2, ..., n / 2$$$. Similarly we will only reverse rows $$$1, 2, ..., m / 2$$$.

Lets observe cells $$$a[i][j], a[i][m - j + 1], a[n - i + 1][j], a[n - i + 1][m - j + 1]$$$, marked as $$$1, 2, 3, 4$$$ in the following picture.

$$$1, 2, 3, 4$$$ will move only among themselves.

Reversing row/column $$$i$$$ twice $$$<=>$$$ not reversing it at all. Thus we will reverse a row/column once or not reverse it at all.

Lets observe all pairs $$$(i, j), i <= n / 2, j <= m / 2$$$. Lets look at 4 options of reversing/not reversing i/j(e.g. reversing $$$i$$$, not reversing $$$j$$$). Only options that put $$$a[i][j], a[i][n - j + 1], a[n - i + 1][j], a[n - i + 1][n - j + 1]$$$ in their right positions are acceptable. Observe different cases and you will see that only following scenarios are possible.(state of row/column $$$i$$$ shows whether $$$i$$$ is reversed or not)
1) Row i must be reversed/not reversed. We can determine state of row i.
2) Column j must be reversed/not reversed. We can determine state of column j.
3) i_is_reversed must be equal to j_is_reversed. Lets connect row $$$i$$$ and column $$$j$$$ with an unordered edge of first type.
4) i_is_reversed can't be equal to j_is_reversed. Lets connect row $$$i$$$ and column $$$j$$$ with an unordered edge of second type.

If state of $$$i$$$ is already set, $$$i$$$ and $$$j$$$ are connected with an edge than we can determine state of $$$j$$$. Lets iterate over rows $$$1, 2, ..., n / 2$$$ and columns $$$1, 2, ..., m / 2$$$. If current row's/column's state is already determined than lets throw a DFS from current row/column, setting other rows/columns that DFS gets to. If current row's/column's state is not determined than lets try both options. Set its state to true, send a DFS. Then set its state to false, send a DFS. Choose option with less reverse operations.