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Discuss this article in the forums Are you familiar with the following situation?
In order to avoid this, on each step we can just modify the matching from the previous step, which only takes O(n2) operations.
It’s easy to see that no more than n2 iterations will occur, because every time at least one edge becomes 0-weight. This is simply a function (for each vertex we assign some number called a label).
Then, after the contest, you find out in the editorial that this problem can be simply reduced to a classical one.
If yes, then this tutorial will surely be useful for you. We can also rephrase this problem in terms of graph theory.
(For example, (W4, J4, W3, J3, W2, J2) and (W4, J1, W1) are alternating paths)If the first and last vertices in alternating path are exposed, it is called (because we can increment the size of the matching by inverting edges along this path, therefore matching unmatched edges and vice versa). And now let’s illustrate these steps by considering an example and writing some code.