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Home/ Questions/Q 466007
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Editorial Team
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Editorial Team
Asked: May 12, 20262026-05-12T23:26:07+00:00 2026-05-12T23:26:07+00:00

I’m currently finding the longest path in a directed acyclic positive-weighted graph by negating

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I’m currently finding the longest path in a directed acyclic positive-weighted graph by negating all edge weights and running Bellman-Ford algorithm. This is working great.

However, I’d like to print the trace of which nodes/edges were used. How can I do that?

The program takes as input the number of nodes, a source, destination and edge weight. Input halts on -1 -1 -1. My code is as follows:

import java.util.Arrays;
import java.util.Vector;
import java.util.Scanner;

public class BellmanFord {
    public static int INF = Integer.MAX_VALUE;

    // this class represents an edge between two nodes
    static class Edge {
        int source; // source node
        int destination; // destination node
        int weight; // weight of the edge
        public Edge() {}; // default constructor
        public Edge(int s, int d, int w) { source = s; destination = d; weight = (w*(-1)); }
    }

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int inputgood = 1;
        int tail;
        int head;
        int weight;
        int count = -1;
        Vector<Edge> edges = new Vector<Edge>(); // data structure to hold graph
        int nnodes = input.nextInt();
        while (inputgood == 1) {
            tail = input.nextInt();
            head = input.nextInt();
            weight = input.nextInt();

            if (tail != -1) {
                edges.add(new Edge(tail, head, weight));
                count++;
            }
            if (tail == -1)
                inputgood = 0;
        }
        int start = edges.get(0).source;
        bellmanFord(edges, nnodes, start);
    }

    public static void bellmanFord(Vector<Edge> edges, int nnodes, int source) {
        // the 'distance' array contains the distances from the main source to all other nodes
        int[] distance = new int[nnodes];
        // at the start - all distances are initiated to infinity
        Arrays.fill(distance, INF);
        // the distance from the main source to itself is 0
        distance[source] = 0;
        // in the next loop we run the relaxation 'nnodes' times to ensure that
        // we have found new distances for ALL nodes
        for (int i = 0; i < nnodes; ++i)
            // relax every edge in 'edges'
            for (int j = 0; j < edges.size(); ++j) {
                // analyze the current edge (SOURCE == edges.get(j).source, DESTINATION == edges.get(j).destination):
                // if the distance to the SOURCE node is equal to INF then there's no shorter path from our main source to DESTINATION through SOURCE
                if (distance[edges.get(j).source] == INF) continue;
                // newDistance represents the distance from our main source to DESTINATION through SOURCE (i.e. using current edge - 'edges.get(j)')
                int newDistance = distance[edges.get(j).source] + edges.get(j).weight;
                // if the newDistance is less than previous longest distance from our main source to DESTINATION
                // then record that new longest distance from the main source to DESTINATION
                if (newDistance < distance[edges.get(j).destination])
                    distance[edges.get(j).destination] = newDistance;
            }
        // next loop analyzes the graph for cycles
        for (int i = 0; i < edges.size(); ++i)
            // 'if (distance[edges.get(i).source] != INF)' means:
            // "
            //    if the distance from the main source node to the DESTINATION node is equal to infinity then there's no path between them
            // "
            // 'if (distance[edges.get(i).destination] > distance[edges.get(i).source] + edges.get(i).weight)' says that there's a negative edge weight cycle in the graph
            if (distance[edges.get(i).source] != INF && distance[edges.get(i).destination] > distance[edges.get(i).source] + edges.get(i).weight) {
                System.out.println("Cycles detected!");
                return;
            }
        // this loop outputs the distances from the main source node to all other nodes of the graph
        for (int i = 0; i < distance.length; ++i)
            if (distance[i] == INF)
                System.out.println("There's no path between " + source + " and " + i);
            else
                System.out.println("The Longest distance between nodes " + source + " and " + i + " is " + distance[i]);
    }
}
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  1. Editorial Team
    Editorial Team
    2026-05-12T23:26:07+00:00Added an answer on May 12, 2026 at 11:26 pm

    You need to slightly modify what you do in the Bellman Ford implementation:

    ...
    int[] lastNode = new int[nnodes];
    lastNode[source] = source;
    for (int i = 0; i < nnodes; ++i)
        for (int j = 0; j < edges.size(); ++j) {
            if (distance[edges.get(j).source] == INF) continue;
            int newDistance = distance[edges.get(j).source] + edges.get(j).weight;
            if (newDistance < distance[edges.get(j).destination])
            {
                distance[edges.get(j).destination] = newDistance;
                lastNode[edges.get(j).destination] = edges.get(j).source;
            }
        }
    

    Printing individual paths then becomes:

    static void printPath(int source, int end, int[] lastNodes)
    {
        if(source!=end)
            printPath(source, lastNodes[end], lastNodes);
        System.out.print(end+" ");
    }
    

    Which prints the path in order from source node to end node.

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