I think that Alice can orient the edges by a greedy algorithm so that one can find a subgraph in which every vertex has a positive outdegree (Explanation

). Now take the longest path, and look at its endpoint, call it $v$. Because of the maximality of the choice and the fact that $\mathrm{deg}^{+}(v)\ge 1$, it must have have a directed edge with a previous vertex from the path, hence we have found an oriented cycle and Alice wins.

You have to be careful. If it was given $1\leq m \leq 2012$, then any time Alice orients $\overrightarrow{vw}$, Bob can orient $\overrightarrow{vu}$ for all $u \not \in \{v,w\}$, and eventually he wins. So it's anything but intuitive ... What Bob wants to achieve is an orientation of $G$ such that the vertices may be enumerated $v_1,v_2,\ldots,v_n$ so that an edge $\overrightarrow{v_iv_j}$ exists if and only if $1\leq i < j \leq n$, and we need investigate if Alice can avoid that. There clearly is a relation between $n$ and the maximum of $m$; for $n\geq 4$ and $\max m = 1$ Alice wins, but for $n= 4$ and $\max m = 2$ Bob wins.

I'm sorry, am I missing something?... crazyfehmy wrote: At each move Bob chooses a positive integer number $m,\: 1\leq m \leq 1000$ and after that directs $m$ undirected edges of $G$ Isn't the condition mentioned in the problem $1\le m\le 1000$ and not $2012$ as you mentioned?

Yes, it is (I wrote "If it was given ..."). But I did see no mention of its usage in your attempted proof

Sorry, true indeed.. I was thinking that it was important to be $1000$ because of the fact that $\dfrac{2014}{2}>1000$, which helped my greedy odd-argument somewhat in my mind. I guess I'll have to think for a better solution

A wins. Her strategy: at her turn find the maximal oriented path $v_1\to \cdots \to v_k$ and choose the least $i$ such that the vertex $v_i$ has at least one unoriented edge in the whole graph; let this edge be $v_i x$ oriented as $x \to v_i$. For B to not lose he must complete the whole graph in at most $2013$ moves, but $1001\cdot 2013< 2013\cdot 1007$, the number of edges. The idea is that after each move A, B the largest path increases, and if it's A's turn, the graph formed by $v_1\to \cdots \to v_k$ must be complete.