# Dijkstra with Constraintby Isai Damier, Android Engineer @ Google

```/*********************************************************************
* Author: Isai Damier
* Title:  Shortest Path with Constraint
* Project: geekviewpoint
* Package: algorithm.graph
*
* Statement:
*   In http://www.geekviewpoint.com/java/graph/dijkstra_shortest_path
*   we talked about finding the shortest path from my local airport in
*   Silicon Valley to my parents' airport in Boston, MA. In the
*   excitement over going to see my parents, I forgot a very important
*   detail: the landing fee each airport charges. You see, not all
*   airports charge the same toll. So I can't just blindly pick the
*   shortest path from Silicon Valley to Boston. I must budget for tolls.
*   So here is the real problem I actually needed to solve:
*
*     In addition to knowing the distance between all pairs of airports
*     that my small airplane could reach in one nonstop fight, I also
*     know how much each airport would charge for toll. So given that
*     I only have c dollars for toll, what is the shortest path from
*     my airport to my parents' airport?
*
* Time-complexity: O((V+E) log V)
*   where V is the number of airports (i.e. vertices/nodes) and E is
*   the number of paths connecting all of the airports.
*
* Dijkstra Constrained:
*
*  It is highly recommended that you read
* http://www.geekviewpoint.com/java/graph/dijkstra_shortest_path before
* proceeding.
*
* ... waiting ... waiting ... waiting ... waiting ... waiting ...
*
* Good. So you got a taste of how life can be when money is not an issue.
* Now to my problem: How do I factor in my financial limitations?
*
* - Instead of using a vector (i.e. 1D array) to track minDist, we must
*   now use a matrix (i.e. 2D array), where the rows will represent the
*   distance from s to the present airport and the columns will
*   represent money left for the journey.
*
* Actual Algorithm:
*
*  Note: we are still using s as my local airport and e as my parents'
*        airport in Boston; v can be any airport.
*
*  0] If I am at airport v, then I need to know how I got there and how
*     far v is from s. Therefore, declare array pred{} to track the
*     predecessor of v and array minDist to track the distance of v
*     from s; so that pred[v]=u would mean I got to v from airport u,
*     and minDist[v][90]=100 would mean the distance between s and v is
*     100 miles and I have \$90 left in my budget after paying v.
*
*  1] Since we have not done any computation yet, we will fill pred
*      with -1s and we will fill minDist with infinities. Both meaning
*     undefined.
*
*  2] Since we are starting at s, we can say that the distance from
*     where we are to s is zero and that we have not yet spent any money.
*     Therefore set minDist[s][c]=0.
*
*  3] Create a priorityQueue containing all vertices v, sorted by
*     minDist; so that vertex s will head the queue since it is the
*     only vertex with a minDist of zero while all others are infinity.
*
*  4] While queue is not empty, remove the head vertex as v:
*
*  5] For each vertex x that is adjacent to v and that is still on the
*     queue, if x would not take us over-budget, recalculate the
*     distance from s to x. To know whether x would take us over-budget,
*     get the money left after v, say m, and subtract the cost of x
*     from m: m - cost[x] >= 0. You may want to manually run the
*     algorithm for a small graph to see how it works.
*
*  6] Re-prioritize the queue based on the recalculated values of
*     minDist and repeat thru the while-loop.
*
*  After step 6 the computation is done. We are left with two arrays:
*  pred and minDist. pred tells us the path from s to e, and minDist
*  tells us the distance along the path. I only care about the path so
*  I extract the path from pred. But you can do so much more as you wish.
**********************************************************************/
package algorithm.graph;

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;

public class ConstrainedShortestPath {

public List<Integer> constrainedShortestPath(
Integer[][] weight,
int[] cost,
int money,
int source,
int finalDestination)
{

//auxiliary constants
final int SIZE = weight.length;
final int EVE = -1;//to indicate no predecessor
final int INFINITY = Integer.MAX_VALUE;

//declare and initialize pred to EVE and minDist to INFINITY
Integer[] pred = new Integer[SIZE];
Integer[][] minDist = new Integer[SIZE][money + 1];
Arrays.fill(pred, EVE);
for (Integer[] m : minDist) {
Arrays.fill(m, INFINITY);
}

//set minDist[source][money]=0, as source is 0 distance from itself.
minDist[source][money] = 0;

PriorityQueue<Integer[]> pq = createPriorityQueue(money, minDist);

while (!pq.isEmpty()) {
updatePriorityQueue(pq);
int v = pq.remove()[0];
for (Integer nodeData[] : adjacency(weight, pq, v)) {
Integer x = nodeData[0];
if (null == x) {
continue;
}

//find how much landing at v actually cost.
//If you can't see this, run a small graph manually.
//Recall that the first v is actuall s with all the money intact.
int m = -1;
while (m < money && minDist[v][++m] == Integer.MAX_VALUE);

//triangle inequality
int c = m - cost[x];
if (c >= 0 && minDist[x][c] >(long) minDist[v][m]+weight[v][x]){
minDist[x][c] = minDist[v][m] + weight[v][x];
nodeData[1] = minDist[x][c];
pred[x] = v;
}
}
}

List<Integer> result = new ArrayList<Integer>();
int p = finalDestination;
while (-1 != pred[p]) {
p = pred[p];
}
Collections.reverse(result);
return result;
}//

private PriorityQueue<Integer[]> createPriorityQueue(
int money, Integer[][] minDist)
{
PriorityQueue<Integer[]> pq = new PriorityQueue<Integer[]>(11,
new Comparator<Integer[]>() {
public int compare(Integer[] A, Integer[] B) {
return A[0] < B[0] ? -1 : 1;
}
});
for (int v = 0; v < minDist.length; v++) {
}
return pq;
}// createPriorityQueue

/******************************************************************
*  Retrieve all the neighbors of vertex v that are
*  in the priority queue pq.
*****************************************************************/
PriorityQueue<Integer[]> pq, int v) {
List<Integer[]> result = new ArrayList<Integer[]>();
for (Integer[] ent : pq) {// {u,key[u]}
int u = ent[0];
if (G[v][u] != null) {
}
}
return result;
}// pqNeighbors

/*****************************************************************
* Re-prioritize the queue based on changes in the
*    minDist array.
*
* Technical Details: Dijktra's algorithm requires a priority queue
*    that changes continuously to reflect changes in minDist.
*    For Java it does not suffice to simply pass new values to
*    the array objects that constitute the queue. The
*    PriorityQueue data structure in Java only checks its structure
*    when it is adding or removing elements. It is unaware of any
*    direct changes to the objects it comprises. Therefore to force
*    the queue to re-prioritize, an element is removed and then
*
*****************************************************************/
private void updatePriorityQueue(PriorityQueue<Integer[]> pq) {
}
}// ConstrainedShortestPath
```
```package algorithm.graph;

import static org.junit.Assert.*;

import java.util.Arrays;
import java.util.List;

import org.junit.After;
import org.junit.Before;
import org.junit.Test;

public class ConstrainedShortestPathTest {

ConstrainedShortestPath dijkstra;

@Before
public void setUp() throws Exception {
dijkstra = new ConstrainedShortestPath();
}

@After
public void tearDown() throws Exception {
dijkstra = null;
}

@Test
public void testConstrainedShortestPath() {
Integer[][] D = {
{null, 2, 2, null, null, null, null, null, null, null},
{null, null, null, 4, null, 16, null, null, null, null},
{null, null, null, null, 3, null, null, 8, null, null},
{null, null, null, null, null, null, 4, 4, null, null},
{null, null, null, null, null, null, null, 4, null, null},
{null, null, null, null, null, null, null, null, 13, null},
{null, null, null, null, null, null, null, null, null, 2},
{null, null, null, null, null, null, null, null, null, 6},
{null, null, null, null, null, null, null, null, null, 4},
{null, null, null, null, null, null, null, null, 5, null}};
int[] S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9};
int money = 199;
List<Integer> expected = Arrays.asList(0, 1, 5, 8);
List<Integer> result =
dijkstra.constrainedShortestPath(D, S, money, 0, 8);

assertEquals(expected,result);
}
}```