After some thinking, I found an answer. If one has a better one I'll accept it.

From a cost matrix of shape $n\times m$ with $n<m$, it is easy to add nodes that will not change anything by giving all their incident edges the weight 0, that is adding $(m-n)*m$ edges with weight 0.

After some thinking, I found an answer. If one has a better one I'll accept it.

From a cost matrix of shape $n\times m$ with $n<m$, it is easy to add nodes that will not change anything by giving all their incident edges the same weight $w$, that is adding $(m-n)*m$ edges.

After some thinking, I found an answer. If one has a better one I'll accept it.

From a cost matrix of shape $n\times m$ with $n<m$, it is easy to add nodes that will never be taken by giving them an infinite weight, that is adding $m-n$ lines with infinite weight. Note that one may have to modify the algorithm used to find the bipartite matching to support infinite weights.

After some thinking, I found an answer. If one has a better one I'll accept it.

From a cost matrix of shape $n\times m$ with $n<m$, it is easy to add nodes that will not change anything by giving all their incident edges the weight 0, that is adding $(m-n)*m$ edges with weight 0.