The optimal solution described here got WA in cses. https://usaco.guide/solutions/cses-1707?lang=cpp
The early return (GOTO) misses some shorter cycles. for graphs such as:
6 7
1 2
2 3
3 4
4 1
1 6
1 5
5 6
We can modify the algorithm above to return either the length of the shortest cycle or the length of the shortest cycle plus one in \mathcal{O}(N^2) time.
It’s not a solution to the CSES problem.
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