POJ 3013 - Big Christmas Tree
Problem : Raw shortest path.

POJ 3463 - Sightseeing
Problem : The number of shortest path and the path larger than shortest path by 1.
Solution : The path P is larger than shortest path by 1 if and only if there is only one edge (u->v) in P, s.t. P(s->u) is a subset of shortest path from s to t, P(v->t) is the shortest path from v to t.
Proof :
1. If there is no such edge in P, there is another path P' which is shorter than P but larger than shortest path.
2. If there are multiple such edge in P, P must be shortest path.

POJ 3621 - Sightseeing Cows
Problem : Find a cycle, to make sum(value(v))/sum(value(e)) maximum.
Solution : 1. The result cycle must contain each vertice only once. 2. Thus use parameter searching method.
Proof : If the cycle contains a vertice more than once, it can be divided to two or more cycles. Thus one of the sub-cycle must have the no worse answer.