Solving the SPOJ Problem Using Bipartite Graph Matching and Auction Algorithm

Introduction

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The SPOJ problem described here involves a specific challenge in algorithm design related to the allocation of hobbits to holes based on certain constraints. This problem can be effectively addressed by modeling it using concepts from graph theory, specifically bipartite graphs and matching problems. In this article, we explore the solution approach outlined, including the rationale behind the bipartite graph model and the auction algorithm for finding the optimal matching.

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Problem Statement

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The problem can be simplified as follows: given a set of hobbits and holes, each hobbit needs to be matched to a hole such that the total cost is minimized, with the cost of secondary holes increased by C units. The goal is to find K such matches with the minimal cost of the most expensive edge in the match.

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Modeling the Problem

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To solve this, the problem is modeled using a bipartite graph:

r r r Nodes in the Left Layer (Hobbits): Each hobbit represents a node in the left layer of the graph.r Nodes in the Right Layer (Holes): Each hole is represented by two nodes in the right layer - a primary and a secondary node.r Edges with Costs: There is an edge between every hobbit and each primary hole with a cost equivalent to the time it takes for the hobbit to reach the hole. Additionally, there is an edge between every hobbit and the secondary hole with a cost that is C units more than the time to the primary hole.r r r

This modeling ensures that if two hobbits need to hide in the same hole, they can prepare it simultaneously, reducing the total time cost.

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Rationale of Modeling

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Let's discuss the correctness of this model:

r r r Single Hobbit per Hole: If only one hobbit should take a hole, it can be modeled by choosing the cheaper primary hole.r Two Hobbits per Hole: If two hobbits need to hide in the same hole, they can run to the hole as fast as they can, and once the first hobbit reaches the hole, it starts preparing it for the second hobbit. The best strategy is to pick the cheaper of the two edges leading to the secondary hole plus the edge to the primary hole to minimize the total time.r r r

The inverse is also true: every matching in the graph, where all edges are not more expensive than t, can be translated into a story where hobbits matched to secondary holes dig shelters for hobbits matched to primary holes, ending in t seconds.

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Solving the Problem Using Auction Algorithm

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The solution can be found using an auction algorithm that iteratively determines the maximum matching in the graph. Here’s the detailed process:

r r r Initialization: Start with an empty graph and define the source and target nodes.r Sorting Edges: Sort the edges by cost (ascending order).r Processing Edges: Add edges one by one, starting from the cheapest to the most expensive, and update the maximum matching.r Residual Graph: Use a directed residual graph to represent the matching. The direction of each edge changes based on whether it is in the current matching or not. An edge leads from the source to a hobbit if the hobbit is not matched, and to the target if the hole is not taken.r Augmenting Paths: Use Breadth-First Search (BFS) or Depth-First Search (DFS) to find augmenting paths in the residual graph and update the matching. If no more paths can be found, increase t and add more edges.r Termination: When K augmenting paths are found, the solution is complete.r r r

The runtime complexity is O(nm log n) for sorting plus O(nmk). Using a heap instead of sorting can potentially reduce the complexity, and further optimization could be achieved with Dinitz's algorithm, though this would increase the complexity of the solution.

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Conclusion

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The problem described can be effectively solved using the bipartite graph matching approach and the auction algorithm. This method ensures that the solution is optimal in terms of minimizing the most expensive edge in the matching. By carefully modeling the problem and using efficient algorithms, we can achieve a practical and efficient solution to the SPOJ problem.

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