The problem using XOR is a red herring, and the intended solution does not depend
on excitement of a game being computed via XOR.

With the exception of the strongest player, each player loses exactly one game.
Therefore, we can match every loser to the player they lose to. This lets us
model the tournament as a min-cost max-flow problem. Each player has two vertices
associated with them, a "loser" vertex and a "winner" vertex. We connect each loser
to a winner if a match between those two would result in the "winner" winning, with capacity
1 and cost equal to the excitement of the game. Losers are connected to the source with
capacity 1 and cost 0, and winners are connected to the sink with capacity $g-1$ and cost
0, with the exception of the strongest player who cannot lose and is therefore connected
to the sink with capacity $g$.

The maximum variant can be solved in similar fashion by negating all the costs
and then negating the final cost when running flow again.