/// <summary>
/// check Digraph is valid
/// </summary>
/// <param name="g"></param>
/// <returns></returns>
public bool IsValid(EdgeWeightedDigraph g)
{
if (g == null || g.V <= 0)
{
return(false);
}
foreach (DirectedEdge e in g.Edges())
{
if (e.Weight < 0)
{
return(false);
}
}
//TODO: add other check here
return(true);
}
/// <summary>
/// constructor
/// for any vertex v reachable from s, the path must at most v-1 edges.
/// we can get (vi-1,vi) in sequence because if p = (v0,v1,...,vk) is shortest path from s =v0 to vk, we relax the edges of p in order (v0,v1) (v1,v2) (vk-1,vk), then we can get the distTo
/// <param name="g"></param>
/// <param name="source"></param>
public BellmanFordAlgorithm(EdgeWeightedDigraph g, int source) : base(g, source)
{
List <DirectedEdge> edges = g.Edges();
for (int i = 0; i < g.V - 1; i++)
{
foreach (DirectedEdge e in edges)
{
base.Relax(e);
}
}
//assume this is negative cycle (v0,v1,...vk) vk=v0. then there must be at least one distTo[e.To] > distTo[e.From] + e.Weight
if (edges.Any(e => distTo[e.To] > distTo[e.From] + e.Weight))
{
IsNegativeCycle = true;
throw new Exception("there is negative cycle");
}
}
