public RoadTripMagnitudeResult GetRoadTripMagnitude(Node node)
{
RoadTripMagnitudeResult result;
if (Magnitudes.TryGetValue(node, out result))
{
return(result);
}
result = new RoadTripMagnitudeResult()
{
Magnitude = 0
};
foreach (var neighbor in node.Neighbors)
{
int neighborPopulation = GetPopulation(neighbor, node);
if (neighborPopulation > result.Magnitude)
{
result.Node = neighbor;
result.Magnitude = neighborPopulation;
}
}
Magnitudes.Add(node, result);
return(result);
}
static void Main(string[] args)
{
// Given a bi-directionally linked graph of nodes, where each node represents
// a location with a population, consider the case where the entire population
// must travel on a road trip to visit a given node. Nodes must travel through
// their neighbors to get to the ultimate destination. For any given node, find
// which neighbor will be bringing the largest population, and the magnitude
// that neighbor is bringing. There are no cycles in the graph.
//
// Example: In the below example, if everyone needs to travel to Denver, then
// 6 people are coming from Cheyenne, 15 are coming from Santa Fe (the 7 in
// Santa Fe itself, plush 3 from Las Cruce and 5 from Albuquerque), and 11 are
// coming from Topeka. So the answer would be (Santa Fe, 15) because that is the
// neighbor delivering the largest magnitude.
//
// Similarly, if everyone needs to travel to Santa Fe, then the answer would be
// (Denver, 29).
//
// Assume the graph is large and the question may be posed repeatedly (i.e., think
// about ways to maximize performance).
var graph = new Graph();
var denver = new Node()
{
Name = "Denver", Population = 12
};
var cheyenne = new Node {
Name = "Cheyenne", Population = 6
};
var santafe = new Node()
{
Name = "Santa Fe", Population = 7
};
var lascruces = new Node {
Name = "Las Cruces", Population = 3
};
var albuquerque = new Node()
{
Name = "Albuquerque", Population = 5
};
var topeka = new Node()
{
Name = "Topeka", Population = 8
};
var columbia = new Node()
{
Name = "Columbia", Population = 1
};
var beatrice = new Node()
{
Name = "Beatrice", Population = 2
};
denver.Neighbors.Add(cheyenne);
denver.Neighbors.Add(santafe);
denver.Neighbors.Add(topeka);
cheyenne.Neighbors.Add(denver);
santafe.Neighbors.Add(denver);
santafe.Neighbors.Add(lascruces);
santafe.Neighbors.Add(albuquerque);
lascruces.Neighbors.Add(santafe);
albuquerque.Neighbors.Add(santafe);
topeka.Neighbors.Add(denver);
topeka.Neighbors.Add(columbia);
topeka.Neighbors.Add(beatrice);
columbia.Neighbors.Add(topeka);
beatrice.Neighbors.Add(topeka);
graph.Nodes.Add(denver);
graph.Nodes.Add(cheyenne);
graph.Nodes.Add(santafe);
graph.Nodes.Add(lascruces);
graph.Nodes.Add(albuquerque);
graph.Nodes.Add(topeka);
graph.Nodes.Add(columbia);
graph.Nodes.Add(beatrice);
RoadTripMagnitudeResult result = null;
Console.WriteLine("Travel to Denver");
result = graph.GetRoadTripMagnitude(denver);
Console.WriteLine(result.Node.Name);
Console.WriteLine(result.Magnitude);
Console.WriteLine();
Console.WriteLine("Travel to Cheyenne");
result = graph.GetRoadTripMagnitude(cheyenne);
Console.WriteLine(result.Node.Name);
Console.WriteLine(result.Magnitude);
Console.WriteLine();
Console.WriteLine("Travel to Santa Fe");
result = graph.GetRoadTripMagnitude(santafe);
Console.WriteLine(result.Node.Name);
Console.WriteLine(result.Magnitude);
Console.WriteLine();
Console.WriteLine("Travel to Topeka");
result = graph.GetRoadTripMagnitude(topeka);
Console.WriteLine(result.Node.Name);
Console.WriteLine(result.Magnitude);
Console.WriteLine();
}
