shortest path to x with weighted graphs
shortest path to x with unweighted graph use breadth-first search
only works for DAGs. cannot have negative weighted edges (use Bellman-Ford algorithm) because a node cannot be reprocessed
table of the cost (how expensive it is to get to there) of each node
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class DijkstrasAlgorithm {
// the graph
private static Map<String, Map<String, Double>> graph = new HashMap<>();
private static List<String> processed = new ArrayList<>();
private static String findLowestCostNode(Map<String, Double> costs) {
Double lowestCost = Double.POSITIVE_INFINITY;
String lowestCostNode = null;
// Go through each node
for (Map.Entry<String, Double> node : costs.entrySet()) {
Double cost = node.getValue();
// If it's the lowest cost so far and hasn't been processed yet...
if (cost < lowestCost && !processed.contains(node.getKey())) {
// ... set it as the new lowest-cost node.
lowestCost = cost;
lowestCostNode = node.getKey();
}
}
return lowestCostNode;
}
public static void main(String[] args) {
graph.put("start", new HashMap<>());
graph.get("start").put("a", 6.0);
graph.get("start").put("b", 2.0);
graph.put("a", new HashMap<>());
graph.get("a").put("fin", 1.0);
graph.put("b", new HashMap<>());
graph.get("b").put("a", 3.0);
graph.get("b").put("fin", 5.0);
graph.put("fin", new HashMap<>());
// The costs table
Map<String, Double> costs = new HashMap<>();
costs.put("a", 6.0);
costs.put("b", 2.0);
costs.put("fin", Double.POSITIVE_INFINITY);
// the parents table
Map<String, String> parents = new HashMap<>();
parents.put("a", "start");
parents.put("b", "start");
parents.put("fin", null);
String node = findLowestCostNode(costs);
while (node != null) {
Double cost = costs.get(node);
// Go through all the neighbors of this node
Map<String, Double> neighbors = graph.get(node);
for (String n : neighbors.keySet()) {
double newCost = cost + neighbors.get(n);
// If it's cheaper to get to this neighbor by going through this node
if (costs.get(n) > newCost) {
// ... update the cost for this node
costs.put(n, newCost);
// This node becomes the new parent for this neighbor.
parents.put(n, node);
}
}
// Mark the node as processed
processed.add(node);
// Find the next node to process, and loop
node = findLowestCostNode(costs);
}
System.out.println("Cost from the start to each node:");
System.out.println(costs); // { a: 5, b: 2, fin: 6 }
}
}
# the graph
graph = {}
graph["start"] = {}
graph["start"]["a"] = 6
graph["start"]["b"] = 2
graph["a"] = {}
graph["a"]["fin"] = 1
graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5
graph["fin"] = {}
# the costs table
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity
# the parents table
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None
processed = []
def find_lowest_cost_node(costs):
lowest_cost = float("inf")
lowest_cost_node = None
# Go through each node.
for node in costs:
cost = costs[node]
# If it's the lowest cost so far and hasn't been processed yet...
if cost < lowest_cost and node not in processed:
# ... set it as the new lowest-cost node.
lowest_cost = cost
lowest_cost_node = node
return lowest_cost_node
# Find the lowest-cost node that you haven't processed yet.
node = find_lowest_cost_node(costs)
# If you've processed all the nodes, this while loop is done.
while node is not None:
cost = costs[node]
# Go through all the neighbors of this node.
neighbors = graph[node]
for n in neighbors.keys():
new_cost = cost + neighbors[n]
# If it's cheaper to get to this neighbor by going through this node...
if costs[n] > new_cost:
# ... update the cost for this node.
costs[n] = new_cost
# This node becomes the new parent for this neighbor.
parents[n] = node
# Mark the node as processed.
processed.append(node)
# Find the next node to process, and loop.
node = find_lowest_cost_node(costs)
print "Cost from the start to each node:"
print costs