Question

Can someone please help with this task for my current code?

Task 3 (50 pts (Extra Credit)). Implement a function to organize the shortest path for each node as a string. For example, if a node 4’s shortest path is 0→2→1→4, you can generate a string variable s=“0→2→1→4”. Modify the displaydistance() function to show the shortest distance and the shortest path for each node.

– Hint 1:You definitely need to do operation for the π variable in this task. Feel free to add any class member data based on your need.

Here is my current code:

public class Graph3 {
   int n;
   int[][] A;
   int[] d;   //shortest distance
   /**
   * @param args
   */
  
   public Graph3 () {
      
   }
  
   public Graph3 (int _n, int[][] _A) {
       n = _n;
       A = _A;
       d = new int[n];
   }
  
   public void initialize_single_source(int s) {
       for (int i = 0; i < n; i++)
           d[i] = Integer.MAX_VALUE;
       d[s] = 0;
   }
  
   public void relax (int u, int v) {
       if (d[v] > (d[u] + A[u][v]))
           d[v] = d[u] + A[u][v];
   }
  
   public boolean bellman_ford (int s) {
       //fill in your program
initialize_single_source(s);
int vertices = A.length;
for (int i=1; i<=vertices-1; i++) {
for (int u=0; u<=vertices-1; u++) {
for (int v=0; v<=vertices-1; v++) {
if (A[u][v] != 0) {
relax (u, v);
}
}
}
}
for (int u=0; u<=vertices-1;u++) {
for (int v=0; v<=vertices-1; v++) {
if (A[u][v] != 0) {
if (d[v]>d[u]+A[u][v]) {
return false;
}
}
}
}
return true;
   }
  
   public void dijkstra (int s) {
       //fill in your program
initialize_single_source(0);
int vertices = A.length;
int visit[] = new int[vertices];
for (int i=0; i<vertices; i++) {
visit[i] = 0;
}
for (int count=0; count<vertices-1; count++) {
int min = Integer.MAX_VALUE, u=0;
for (int v=0; v<vertices; v++) {
if (visit[v] == 0 && d[v] <= min) {
min = d[v];
u=v;
}
}
visit[u] = 1;
for (int v=0; v<vertices; v++) {
if (visit[v]==0 && A[u][v]>0 && d[u] != Integer.MAX_VALUE && d[u]+A[u][v]<d[v]) {
d[v] = d[u] + A[u][v];
}
}
}
   }
  
   public void display_distance () {
       for (int i = 0; i < n; i++) {
           System.out.println(i + ": " + d[i]);
}
   }

  
   public static void main(String[] args) {
       // TODO Auto-generated method stub
       int n = 5;
       int[][] A = {
       {0, 6, 0, 7, 0},
       {0, 0, 5, 8, -4},
       {0, -2, 0, 0, 0},
       {0, 0, -3, 0, 9},
       {2, 0, 7, 0, 0}
       };
       Graph3 g1 = new Graph3(n, A);
       g1.bellman_ford(0);
       g1.display_distance();
      
       System.out.println("-----------------------");
      
       int[][] B = {
       {0, 10, 0, 5, 0},
       {0, 0, 1, 2, 0},
       {0, 0, 0, 0, 4},
       {0, 3, 9, 0, 2},
       {7, 0, 6, 0, 0}
       };
       Graph3 g2 = new Graph3(n, B);
       g2.dijkstra(0);
       g2.display_distance();
   }

}

Answer 1

Can someone please help with this task for my current code?
Task 3 (50 pts (Extra Credit)). Implement a function to organize the shortest path for each node as a string. For example, if a node 4’s shortest path is 0→2→1→4, you can generate a string variable s=“0→2→1→4”. Modify the displaydistance() function to show the shortest distance and the shortest path for each node.
– Hint 1:You definitely need to do operation for the π variable in this task. Feel free to add any class member data based on your need.
Here is my current code:
public class Graph3 {
int n;
int[][] A;
char s="";
int[] d; //shortest distance
/**
* @param args
*/
  
public Graph3 () {
  
}
  
public Graph3 (int _n, int[][] _A) {
n = _n;
A = _A;
d = new int[n];
}
  
public void initialize_single_source(int s) {
for (int i = 0; i < n; i++)
d[i] = Integer.MAX_VALUE;
d[s] = 0;
}
  
public void relax (int u, int v) {
if (d[v] > (d[u] + A[u][v]))
d[v] = d[u] + A[u][v];
}
  
public boolean bellman_ford (int s) {
//fill in your program
int[] prev = new int[vertices];
  
initialize_single_source(s);
int vertices = A.length;
for (int i=1; i<=vertices-1; i++) {
for (int u=0; u<=vertices-1; u++) {
for (int v=0; v<=vertices-1; v++) {
if (A[u][v] != 0) {
relax (u, v);
}
}
}
}
for (int u=0; u<=vertices-1;u++) {
for (int v=0; v<=vertices-1; v++) {
if (A[u][v] != 0) {
if (d[v]>d[u]+A[u][v]) {
return false;
}
}
}
}
return true;
}
  
public void dijkstra (int s) {
//fill in your program
initialize_single_source(0);
int vertices = A.length;
int visit[] = new int[vertices];
for (int i=0; i<vertices; i++) {
visit[i] = 0;
}
for (int count=0; count<vertices-1; count++) {
int min = Integer.MAX_VALUE, u=0;
for (int v=0; v<vertices; v++) {
if (visit[v] == 0 && d[v] <= min) {
min = d[v];
u=v;

}
}
visit[u] = 1;
for (int v=0; v<vertices; v++) {
if (visit[v]==0 && A[u][v]>0 && d[u] != Integer.MAX_VALUE && d[u]+A[u][v]<d[v]) {
d[v] = d[u] + A[u][v];
}
}
}
}
  
public void display_distance () {
for (int i = 0; i < n; i++) {
System.out.println(i + ": " + d[i]);
}
}

  
public static void main(String[] args) {
// TODO Auto-generated method stub
int n = 5;
int[][] A = {
{0, 6, 0, 7, 0},
{0, 0, 5, 8, -4},
{0, -2, 0, 0, 0},
{0, 0, -3, 0, 9},
{2, 0, 7, 0, 0}
};
Graph3 g1 = new Graph3(n, A);
g1.bellman_ford(0);
g1.display_distance();
  
System.out.println("-----------------------");
  
int[][] B = {
{0, 10, 0, 5, 0},
{0, 0, 1, 2, 0},
{0, 0, 0, 0, 4},
{0, 3, 9, 0, 2},
{7, 0, 6, 0, 0}
};
Graph3 g2 = new Graph3(n, B);
g2.dijkstra(0);
g2.display_distance();
}
}