*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.

((W4, J4, W3, J3, W2, J2) – augmenting alternating path)A tree which has a root in some exposed vertex, and a property that every path starting in the root is alternating, is called an . Find(1)and replace existing labeling with the next one:(2)Now replace with Step 3. As an example we’ll use the previous one, but first let’s transform it to the maximum-weighted matching problem, using the second method from the two described above.For each vertex from left part (workers) find the minimal outgoing edge and subtract its weight from all weights connected with this vertex. Actually, this step is not necessary, but it decreases the number of main cycle iterations. Find the maximum matching using only 0-weight edges (for this purpose you can use max-flow algorithm, augmenting path algorithm, etc.). Step 2) Let and adjust the weights using the following rule: Step 3) Repeat Step 1 until solved.But there is a nuance here; finding the maximum matching in step 1 on each iteration will cause the algorithm to become O(n5). Wiley Online Library requires cookies for authentication and use of other site features; therefore, cookies must be enabled to browse the site.Detailed information on how Wiley uses cookies can be found in our Privacy Policy.There are also implementations of Hungarian algorithm that do not use graph theory.Rather, they just operate with cost matrix, making different transformation of it (see [1] for clear explanation).We’ll handle the assignment problem with the Hungarian algorithm (or Kuhn-Munkres algorithm).I’ll illustrate two different implementations of this algorithm, both graph theoretic, one easy and fast to implement with O(n4) complexity, and the other one with O(n3) complexity, but harder to implement.We’ll not touch these approaches, because it’s less practical for Top Coder needs.As mentioned above, we are dealing with a bipartite graph.

## Comments Transportation Problem And Assignment Problem

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