UVa 10735 - Euler Circuit

10735 - Euler Circuit

 * http://acm.uva.es/p/v107/10735.html

Summary
Given a graph $$G$$, which contains both directed edges and undirected edges, find a closed path in it, in which each edge is included exactly once.

Explanation
Recall, when Euler tour exists in a directed graph: the underlying undirected graph is connected, and the in-degree of each vertex is equal to the out-degree.

In this problem, some of the graph's edges may be undirected. If we can orient them in such a way, that the in-degree of each vertex will be equal to its out-degree, then the problem will be reduced to finding a tour in a directed graph. Such orientation can be found by solving the following bipartite matching problem.

Construct a bipartite graph $$H$$. In one partition put all edges (both directed and undirected) of $$G$$, and the other partition contains $$G$$'s vertices. For every edge we have to know, which of its two endpoints is the head. So, connect every object (edge of $$G$$) in the first partition with its $$G$$'s endpoints in the second partition.

We'll be finding a matching in this graph. If an undirected edge $$e=(u,v)$$ of $$G$$ will be matched with $$v$$, it means, that in the final directed graph, the edge $$e$$ will go from vertex $$u$$ to vertex $$v$$.

Each matched $$H$$'s edge of $$(e,v)$$ will contribute to the in-degree of vertex $$v$$ in the directed graph, and unmatched edge $$(e,u)$$ contributes to the out-degree of $$u$$.

Since we want to make the in-degree and out-degree of each vertex equal, each vertex must have an equal number of matched and unmatched edges in $$H$$. Additionally, each directed edge has to be matched with its respective head from $$G$$.

After finding a matching in $$H$$, satisfying the outlined constraints, we can assign direction to each undirected $$G$$'s edge and find Euler tour is the resulting directed graph with any standard algorithm. If a matching doesn't exist, there will be no Euler tour in the original graph.

Input
2 6 8 1 3 U 1 4 U 2 4 U 2 5 D 3 4 D 4 5 U 5 6 D 5 6 U 4 4 1 2 D 1 4 D 2 3 U 3 4 U

Output
1 3 4 2 5 6 5 4 1

No euler circuit exist

Implementations

 * Sweepline - C++