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Thoughts.
- At best 10^6 vertices, at worst 10^3 times processing.
- If you check every vertex every time, you won’t make it in 10^9, but you don’t do that.
- If we do a width-first search, the process will be completed in 10^6 orders of magnitude, which is OK.
- Official Explanation OK
- I was unconsciously rewriting the problem.
- For all black squares, if there is a white square adjacent to the black square, it is black” is the same as “For all black squares, if there is a white square adjacent to the black square, it is black
- The white square adjacent to the black square is turned to black
- Sure would have bothered me if I hadn’t noticed this rephrasing.
- In the latter, once a black square has been dealt with, there will be no more white squares around it Never have to deal with it again., and this is the principle of reduced order.
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