visual basic barcode program 10: AI-Based Problem Solving in Java

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10: AI-Based Problem Solving
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System.out.println(to); System.out.println("Distance is " + dist); }
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Notice the use of a second stack called rev. The solution stored in btStack is in reverse order, with the top of the stack holding the last connection and the bottom of the stack holding the first connection. Thus, it must be reversed in order to display the connection in the proper sequence. To put the solution into its proper order, the connections are popped from btStack and pushed onto rev. The entire depth-first search program follows:
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// Find connections using a depth-first search. import java.util.*; import java.io.*; // Flight information. class FlightInfo { String from; String to; int distance; boolean skip; // used in backtracking FlightInfo(String f, String t, int d) { from = f; to = t; distance = d; skip = false; } } class Depth { final int MAX = 100; // maximum number of connections // This array holds the flight information. FlightInfo flights[] = new FlightInfo[MAX]; int numFlights = 0; // number of entries in flight array Stack btStack = new Stack(); // backtrack stack public static void main(String args[]) { String to, from; Depth ob = new Depth(); BufferedReader br = new
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BufferedReader(new InputStreamReader(System.in)); ob.setup(); try { System.out.print("From "); from = br.readLine(); System.out.print("To "); to = br.readLine(); ob.isflight(from, to); if(ob.btStack.size() != 0) ob.route(to); } catch (IOException exc) { System.out.println("Error on input."); } } // Initialize the flight database. void setup() { addFlight("New York", "Chicago", 900); addFlight("Chicago", "Denver", 1000); addFlight("New York", "Toronto", 500); addFlight("New York", "Denver", 1800); addFlight("Toronto", "Calgary", 1700); addFlight("Toronto", "Los Angeles", 2500); addFlight("Toronto", "Chicago", 500); addFlight("Denver", "Urbana", 1000); addFlight("Denver", "Houston", 1000); addFlight("Houston", "Los Angeles", 1500); addFlight("Denver", "Los Angeles", 1000); } // Put flights into the database. void addFlight(String from, String to, int dist) { if(numFlights < MAX) { flights[numFlights] = new FlightInfo(from, to, dist); numFlights++; } else System.out.println("Flight database full.\n");
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10: AI-Based Problem Solving
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} // Show the route and total distance. void route(String to) { Stack rev = new Stack(); int dist = 0; FlightInfo f; int num = btStack.size(); // Reverse the stack to display route. for(int i=0; i < num; i++) rev.push(btStack.pop()); for(int i=0; i < num; i++) { f = (FlightInfo) rev.pop(); System.out.print(f.from + " to "); dist += f.distance; } System.out.println(to); System.out.println("Distance is " + dist); } /* If there is a flight between from and to, return the distance of flight; otherwise, return 0. */ int match(String from, String to) { for(int i=numFlights-1; i > -1; i--) { if(flights[i].from.equals(from) && flights[i].to.equals(to) && !flights[i].skip) { flights[i].skip = true; // prevent reuse return flights[i].distance; } } return 0; // not found } // Given from, find any connection. FlightInfo find(String from) {
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for(int i=0; i < numFlights; i++) { if(flights[i].from.equals(from) && !flights[i].skip) { FlightInfo f = new FlightInfo(flights[i].from, flights[i].to, flights[i].distance); flights[i].skip = true; // prevent reuse return f; } } return null; } // Determine if there is a route between from and to. void isflight(String from, String to) { int dist; FlightInfo f; // See if at destination. dist = match(from, to); if(dist != 0) { btStack.push(new FlightInfo(from, to, dist)); return; } // Try another connection. f = find(from); if(f != null) { btStack.push(new FlightInfo(from, to, f.distance)); isflight(f.to, to); } else if(btStack.size() > 0) { // Backtrack and try another connection. f = (FlightInfo) btStack.pop(); isflight(f.from, f.to); } } }
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10: AI-Based Problem Solving
Notice that main( ) prompts you for both the city of origin and the city of destination. This means that you can use the program to find routes between any two cities. However, the rest of this chapter assumes that New York is the origin and Los Angeles is the destination. When run with New York as the origin and Los Angeles as the destination, the following solution is displayed. As Figure 10-5 shows, this is indeed the first solution that would be found by a depth-first search. It is also a fairly good solution.
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