shortest path with two variables - c++

Question

So I'm trying to use a modified Bellman Ford algorithm to find the shortest path from the starting vertex to the ending vertex but I cannot go over a certain distance. So given a graph with edges:
0 1 100 30
0 4 125 50
1 2 50 250
1 2 150 50
4 2 100 40
2 3 90 60
4 3 125 150
Where the each line represents an edge and the first value is the starting vertex, the second value is the end vertex, the third is cost and the fourth is the distance.
With the code I have now when I try to find the cheapest path from 0 to 3 without going over 140 it yields 0 (default when no path is found) instead of 340 (the cost of the cheapest path). Any suggestions on how to alter my code.
Just gonna copy the code down below because this site is not letting me do anything else.
static void BellmanFord(struct Graph *graph, int source, int ending, int max){
int edges = graph->edgeCount;
int vertices = graph->verticesCount;
int* money = (int*)malloc(sizeof(int) * vertices);
int* distance = (int*)malloc(sizeof(int) * vertices);
for (int I = 0; I < vertices; I++){
distance[I] = INT_MAX;
money[I] = INT_MAX;
}
distance[source] = 0;
money[source] = 0;
for (int I = 1; I <= vertices - 1; ++I){
for int j = 0; j < edges; ++j){
int u = graph->edge[j].Source;
int v = graph->edge[j].Destination;
int Cost = graph->edge[j].cost;
int Duration = graph->edge[j].duration;
if ((money[u] != INT_MAX) && (money[u] + Cost < money[v])){
if (distance[u] + Duration <= max){
money[v] = money[u] + Cost;
distance[v] = distance[u] + Duration;
}
}
}
}
if (money[ending] == INT_MAX) cout << "0" << endl;
else cout << money[ending] << endl;
}
Please help! This is probably not that hard but finals are stressing me out

Answer 1

This problem, known as the "constrained shortest path" problem, is much harder to solve than this. The algorithm you provided does not solve it, it only might catch the solution, only by luck, according to the graph's structure.
When this algorithm is applied on the graph you provide, with max-distance = 140, it fails to find the solution, which is 0-->1-->2-->3 (using the edge 1 2 150 50) with total cost of 340 and a distance of 140.
We can observe the reason of failure by logging the updates to the nodes whenever they are updated, and here is the result:
from node toNode newCost newDistance
0 1 100 30
0 4 125 50
1 2 250 80
4 2 225 90
Here the algorithm is stuck and cannot go further, since any progress from this point will lead to paths that exceed the max distance (of 140). As you see, node 2 has found the path 0-->4--2 which is the lowest-cost from node 0 while respecting the max-distance constraint. But now, any progress from 2 to 3 will exceed the distance of 140 (it will be 150, because 2->3 has a distance of 60.)
Running again with max-distance=150 will find the path 0-->4->3 with cost 315 and distance 150.
from node toNode newCost newDistance
0 1 100 30
0 4 125 50
1 2 250 80
4 2 225 90
2 3 315 150
Obviously this is not the minimum cost path that respects the constraint of distance; the correct should be the same (that it failed to find) in the first example. This again proves the failure of the algorithm; this time it gives a solution but which is not the optimal one.
In conclusion, this is not a programming mistake or bug in the code, it is simply that the algorithm is not adequate to the stated problem.

Answer 2

Okay so right before the
if (money[ending] == INT_MAX) cout << "0" << endl;
I added some code that made it work but I'm wondering will this work for every case or does it need to be altered a little.
if (money[ending] == INT_MAX){
for (int j = 0; j < edges; ++j){
int u = graph->edge[j].Source;
int v = graph->edge[j].Destination;
int Cost = graph->edge[j].cost;
int Duration = graph->edge[j].duration;
if ((distance[u] != INT_MAX) && (distance[u] +Duration < distance[v])){
if (distance[u] + Duration <= max){
money[v] = money[u] + Cost;
distance[v] = distance[u] + Duration;
}
}
}
}