No.
That is, there is no polynomial time algorithm in $m\times n$ that can find a simple cycle with minimal weight in any given weighted complete bipartite directed graph K_$\{m, n\}$ in the sense that it is a NP-complete problem.
To prove it, let us reduce the undirected Hamiltonian cycle problem, one of Karp's 21 NP-complete problems, to the current problem by a polynomial-time reduction.
Let $G$ be an undirected graph with vertices $V$ and edges $E$. For each vertex $u$ in $G$, we create two new vertices, $u_{in}$ and $u_{out}$. Let $V_{in}=\{u_{in}|u\in V\}$. Let $V_{out}=\{u_{out}|u\in V\}$. Let $K$ be the complete bipartite digraph connecting the vertex sets $V_{in}$ and $V_{out}$. For every vertex $u\in G$, both edges $(u_{in},u_{out})$ and $(u_{out}, u_{in})$ in $K$ weigh 0. If edge $(u, v)$ is in $G$, then the both edge $(u_{out},v_{in})$ and edge $(v_{out},u_{in})$ in $K$ weigh $-1$. Any other edge in $K$ weigh $|V|$. $K$ with its edge-weights thus defined is a weighted complete bipartite directed graph with $|V|$ vertices in each of its two vertex sets (that is, $m=n=|V|$ for $K$).
Each simple cycle in $G$ corresponds to a simple cycle in $K$ with the same weight in the natural way as illustrated by the following example. A cycle $a, b, c, a$ in $G$ corresponds to the cycle $a_{in}, a_{out}, b_{in}, b_{out}, c_{in}, c_{out}, a_{in}$ in $K$. In particular, a Hamiltonian Cycle in $G$ corresponds to a simple cycle in $K$ of weight $-|V|$. Conversely, it is easy to see that a simple cycle of weight $-|V|$ in $K$ corresponds to a Hamiltonian Cycle in $G$ under the same correspondence. Note that a simple cycle of weight $-|V|$ in $K$ must be simple cycle of minimal weight in $K$.
Assume we can find a simple cycle $C$ of the minimal weight in $K$ by some algorithm. We can then check the weight of $C$. If the weight of $C$ is $-|V|$, the simple cycle in $G$ that corresponds to $C$ is a Hamiltonian Cycle. Otherwise, there is no Hamiltonian Cycle in $G$. So we have reduced the undirected Hamiltonian cycle problem to the current problem. We can see easily that this reduction is a polynomial-time reduction.
