Shortest cycle for each vertex
An easy way is as follows: For each vertex $s$, perform a BFS visit to discover the shortest cycle containing $s$.

When a visit from $s$ is performed a BFS tree $T_s$ is built and each non-tree edge $(u,v) \not\in E(T)$ for which $u$ and $v$ belong to different subtrees rooted in the children of $s$ in $T$ tree induces a cycle containing $s$.
Since, the shortest cycle containing $s$ must use a non-tree edge $(u,v)$ with this property, it will be found by the visit. You can optionally stop the visit early: if you find a cycle while visiting $u$, you can stop the BFS as soon as all the vertices in the level of $T$ containing of $u$ are visited.

The overall required time is $O(n \cdot m)$.