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Author: sjay05
Subtask 1
Notice that a path from a land cell in row 
~1~ to row 
~N~ is a concatenation of multiple edges each with various lengths. An edge of length 
~l~ exists between two cells of land if both cells are in the same row/column and the space between them on that row/column is occupied by ~l~ puddle cells.
All edges (with a maximum of 
~2 \cdot NM~) can be built by considering rows and columns separately and connecting each land cell to the subsequent cell in the corresponding row/column being iterated.
For ease of storing edges, we map each cell 
~(i, j)~ with a distinct id.
Process the edges in decreasing order by length ~l~ and maintain a bi-directional graph of active edges. After all edges of a certain length ~l~ are added to the graph, we can BFS from all land cells in row ~1~ and track land which cells in row ~N~ are visited (again, using only edges with length 
~\ge l~). Note that there is a maximum of 
~\max(N, M)~ different edge lengths.
Note: To find the cells in row ~N~ that can be reached with a canoe size being infinitely large, we process all edges with 
~l = 0~, and BFS to find the land cells holding this special case.
Time Complexity: 
~\mathcal{O}(NM \cdot \log(NM) + NM \cdot \max(N, M))~
Subtask 2
Using the cell ids, maintain a DSU to query for which cells are connected after a certain number of edges have been processed.
For infinitely large canoe sizes, process the edges with ~l = 0~, adding each to the DSU and checking which land cells in row ~N~ share a set with land cells in row ~1~ in 
~\mathcal{O}(M \log(NM))~ time.
For the remaining edge lengths, process them in decreasing order. After all edges of a certain length ~l~ are added to the DSU, we can repeat the ~\mathcal{O}(M \log(NM))~ check for finding which land cells in row ~N~ can be connected with a land cell in row ~1~ using edge lengths ~\ge l~. Note that there are a maximum of ~\max(N, M)~ different edge lengths.
Time Complexity: 
~\mathcal{O}(NM \cdot \log(NM) + \max(N, M) \cdot M \log(NM))~
Comments
When Testing, I have an alternate solution where I build the graph, and run a variation of dijkstra where I maximize the minimum edge.
Another approach: Build any maximum spanning tree, create a new node connecting first row and run a dfs from it: "minimum edge in the path to a last row cell is the answer".
Complexity:
