Tournament and cycles-reference request I am looking for any reference on the subject of tournaments (simple directed graphs with a unique oriented edge between any two vertices) and cycles. More specifically, given a function $w:E\rightarrow [0,\infty)$ such that for any vertex $v\in V$:
$$ \sum\limits_{(v,u)\in E} w\big( v,u  \big)-\sum\limits_{(u,v)\in E} w\big( u,v  \big)=0 $$
i.e, such that the "out-degree" and "in-degree" is according to weight function and not just the amount of edges "exiting" $v$.

I am wondering whether there is a "classic" result about whether conditions under which one can extend an edge to a cycle. Specifically the edge for which $w$ is maximal.
 A: These sort of weighted tournaments (especially with the extra Eulerian-like condition) are not very commonly studied, but fortunately, cycles in tournaments are easy to find. I'll say what we can say in increasing order of specificity.

In general, for any directed graph, we can partition the vertices into strongly connected components: a vertex set $S$ is a strongly connected component when for any $v,w \in S$ there are directed paths in both directions between $v$ and $w$, and $S$ cannot be made larger and keep this property.
We can replace each strongly connected component by a single vertex to get a graph called the condensation of the original graph. This is always a directed acyclic graph. In particular, if you're looking for cycles in directed graphs, these will always be found within a single strongly connected component and never between two such components. In particular, edges that go from one strongly connected component to another can never be extended to a cycle.
However, within a strongly connected component, every edge $vw$ can be extended to a cycle, because there must be a directed path from $w$ to $v$. 

We haven't used the condition about sums of edge weights so far. We didn't need it to find cycles, but we can use it to make some observations about strongly connected components.
I mentioned that the condensation of a directed graph is acyclic. In particular, it has a source. That source corresponds to a strongly connected component $S$ such that all edges between $S$ and the rest of the graph are oriented away from $S$.
Consider the sum
$$
   \sum_{v \in S} \left(\sum_{u : vu \in E} w(v,u) - \sum_{u : uv \in E} w(u,v)\right).
$$
On the one hand, by the weight condition, the sum is $0$. On the other hand, every edge $uv$ with $u,v \in S$ contributes its weight twice, once with a $+$ sign and once with a $-$ sign, and so they cancel. We are left with only edges with just one endpoint in $S$, and we know all such edges are pointed away from $S$, telling us that
$$
    \sum_{v \in S} \sum_{u \notin S} w(v,u) = 0.
$$
All weights are nonnegative, and so we conclude that all weights on the edges out of $S$ are $0$.
This means we can delete $S$ from the tournament and keep the weight condition true for all remaining vertices. By repeating this argument, we conclude that all edges between two different strongly connected components have weight $0$.
Putting this together with our previous work, we can make some conclusions:


*

*Every edge with positive weight can be extended to a cycle. (It's possible that some of the edges with weight $0$ can be extended as well, just not the ones that go between two strongly connected components.)

*In particular, an edge with the maximum weight can be extended to a cycle, except in the case where all edges have weight $0$ (then, weights tell us nothing).


Actually, all of the above holds for directed graphs and not just tournaments. Since we can delete all edges with weight $0$ without breaking the weight condition, we get the following statement for free:


*

*Each edge with positive weight can be extended to a cycle in which all edge weights are positive.


(This has algorithmic applications: to find the cycle, we can do a depth-first search that only considers edges of positive weight, which helps us because it means we're guaranteed to stay in the same strongly connected component.)
