Can anyone explain implementation details of Tourists(Euler circuits) problem in january long challenge. I couldn’t understand from editorails.
First let’s consider the cases where the answer is going to be “NO”:
- The graph is not connected (more than 1 components)
- There exists vertices (or a single vertex) with odd degree
If either of the above is true, just print NO
otherwise, create an adjacency list for an undirected graph (i.e. for an edge u,v add v to u’s list and vice versa) and use the following algorithm:
for each edge e=(u,v) in E:(vertices adjacent to u) remove v from u's list remove u from v's list add edge u,v to a set or vector (which contains the tour sequence) findtour(v)
and then print the edges in the set or vector you inserted the edges according to the order in the input
for full implementation, check my submission here
implementation details it’s behavior produced by the code which may be relied on by consuming code. Though that behavior not specified by spec the code is written to hence Assignment Moz We will break it to consuming code that’s why it’s bad to rely on them
Here is my simple DFS solution! I just used a reverse edge concept in which i visited the edge and whenever we encounter the reverse edge in DFS, then i just simply visit this by using map<> STL in C++.
In my solution I assigned reverse edge as 1 and directed edge as 0. So in Check() function whenever i visit reverse edge i just visit this by MAP<>.
In printing side whenever i encounter reverse edge, then i simply reverse it. Only! I hope this will be simple from Editorial solution!
If you wanna ask more about this, then freely comment your doubt!
See this code!
@swetankmodi Thanks It really helped me in understanding the solution.
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