// O(|v|+|E|) space to hold all of the nodes for the graph,
// plus 2*|v| for the previous and distances lists required.
// For the heapQ we have O(2*|v|) and for the ArrayQ we have O(|v|),
// as explained in their class descriptions. In the end we also return
// the path which could worst-case cost O(|v|).
// Overall we have O(6*|v| + |E|) in the worst-case.
public static Tuple <List <int>, List <double> > run(
List <PointF> points, List <HashSet <int> > adjacencyList,
int startNode, int stopNode, bool useArray)
{
List <double> distances = new List <double>();
List <int> prev = new List <int>();
for (int i = 0; i < points.Count; i++)
{
distances.Add(int.MaxValue);
prev.Add(-1);
}
distances[startNode] = 0;
IQueue Q;
if (useArray)
{
Q = new ArrayQ(distances);
}
else
{
Q = new HeapQ(distances);
}
int cur;
double dis;
HashSet <int> adjs;
// this is O(|v| * |v|) for the array and O(|v| * 2log(|v|)) for the heap.
// the O(|v|) is common to both because this while loop runs |v| times.
// The O(|v|) for the array is due to the Deletemin, and the O(2log(|v|))
// for the heap is because it has two operations that are O(log(|v|)).
while ((cur = Q.Deletemin(distances)) != -1)
{
adjs = adjacencyList[cur];
foreach (int end in adjs) // runs 3 times
{
dis = distances[cur] + Distance(points[cur], points[end]);
if (dis < distances[end])
{
distances[end] = dis;
prev[end] = cur;
Q.Decreasekey(end, distances); // O(1) for ArrayQ and O(log|v|) for HeapQ
}
}
}
return(new Tuple <List <int>, List <double> >(PathToEnd(startNode, stopNode, prev), distances));
} // end of run()
public static List <int> run(
List <PointF> points, List <HashSet <int> > adjacencyList,
int startNode, int stopNode, bool useArray)
{
List <double> distances = new List <double>();
List <int> prev = new List <int>();
for (int i = 0; i < points.Count; i++)
{
distances.Add(int.MaxValue);
prev.Add(-1);
}
distances[startNode] = 0;
IQueue Q;
if (useArray)
{
Q = new ArrayQ(distances);
}
else
{
Q = new HeapQ(distances);
}
int cur;
double dis;
HashSet <int> adjs;
while ((cur = Q.Deletemin(distances)) != -1)
{
adjs = adjacencyList[cur];
foreach (int end in adjs)
{
dis = distances[cur] + Distance(points[cur], points[end]);
if (dis < distances[end])
{
distances[end] = dis;
prev[end] = cur;
Q.Decreasekey(end, distances);
}
}
}
return(PathToEnd(startNode, stopNode, prev));
} // end of run()
//O(|V|) * (O(OnKeyDecreased) + O(Deletemin))
private void solveButton_Clicked(bool useArray)
{
//***************** Dijkstra's algorithm: *****************//
double[] dist = new double[points.Count];
int[] prev = new int[points.Count];
//O(|V|)
for (int i = 0; i < adjacencyList.Count; i++)
{
dist[i] = Int32.MaxValue;
prev[i] = -1;
}
dist[startNodeIndex] = 0;
IPriorityQ pq;
if (useArray)
{
pq = new ArrayQ();
}
else
{
pq = new HeapQ();
}
pq.Makequeue(dist);
//O(|V|) * (O(OnKeyDecreased) + O(Deletemin))
while (pq.NotEmpty())
{
int u = pq.Deletemin();
//O(3) * O(OnKeyDecreased)
foreach (int v in adjacencyList.ElementAt(u))
{
if (dist[v] > dist[u] + EuclDist(points.ElementAt(u), points.ElementAt(v)))
{
dist[v] = dist[u] + EuclDist(points.ElementAt(u), points.ElementAt(v));
prev[v] = u;
pq.OnKeyDecreased(v);
}
}
}
//****************** End Dijkstra's **********************//
//Get shortest path from prev. Draw it to screen
if (prev[stopNodeIndex] == -1)
{
displayNoPath();
return;
}
List <PointF> path = new List <PointF>();
int curr = stopNodeIndex;
//O(|V|)
while (curr != -1)
{
path.Add(points.ElementAt(curr));
curr = prev[curr];
}
drawPath(path);
labelPath(path);
}
