http://www.spoj.com/problems/NAKANJ/
There are lots of ways to solve this, BFS recursion, DFS dynamic. I suggest using thee BFS (If you need more info on BFS go here) approach, it is simpler to understand but harder to implement. Just keep a queue and push the coordinates of the first cell to it, every time you take out an element check all of the 8 neighbors that cell has and add them to the queue , also don't forget to mark all the visited cells so that you won't have to visit them again and again.Check out the source code for a better understanding.
There are lots of ways to solve this, BFS recursion, DFS dynamic. I suggest using thee BFS (If you need more info on BFS go here) approach, it is simpler to understand but harder to implement. Just keep a queue and push the coordinates of the first cell to it, every time you take out an element check all of the 8 neighbors that cell has and add them to the queue , also don't forget to mark all the visited cells so that you won't have to visit them again and again.Check out the source code for a better understanding.
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