graph 1:  s->a = 2,  s->b = 3,  b->a = -2,  a->c = 1
  vertex   Dijkstra   Bellman-Ford
     s          0             0
     a          2             1
     b          3             3
     c          3             2
  the true shortest path to c is s->b->a->c = 3 + (-2) + 1 = 2;
  Dijkstra froze d[a] = 2 before the edge b->a could improve it.

graph 2:  s->a = 2,  a->b = 1,  b->a = -4   (cycle weight -3)
  Bellman-Ford returns false -- a negative-weight cycle is reachable from s
