Question

2. Apply Dijkstra’s algorithm as discussed in class to solve the single-source shortest-paths problem for the following graph. Consider node a to be the source. (10 points) a. Show the completed table. b. State the shortest path from A to J and state its length. c. State the shortest path from A to K and state its length. d. State the shortest path from A to L and state its length. 3 5 6 4 3 2 1 2. d 5 4 5 3 6 3 h 7 5 6 9 k 00

Answer 1

#include <limits.h>
#include <stdio.h>
#include <stdbool.h>
#define v 12
int min_indx(int key[],bool visited[])
{
int min=INT_MAX,min_idx;
for(int i=0;i<v;i++)
{
if(visited[i]==false && key[i]<=min)
{
min=key[i];
min_idx=i;
}
}
return min_idx;
}

void dijkstra(int graph[v][v])
{
int key[v],u,j;
bool visited[v];
for(int i=0;i<v;i++)
{
visited[i]=false;
key[i]=INT_MAX;
}
key[0]=0;
for(int i=0;i<v-1;i++)
{
u=min_indx(key,visited);
visited[u]=true;
for(int j=0;j<v;j++)
{
if(!visited[j] && graph[u][j] && key[u]!=INT_MAX && key[u]+graph[u][j]<key[j])
{
key[j]=key[u]+graph[u][j];
}
}
}
printf("source \tvertex minimal distance\n");
j=0;
for (char i = 'a'; i < 'l'; i++)
{
printf("%c \t %c \t %d\n",'a', i, key[j++]);
}

}

int main()
{
int graph[v][v] = { { 0, 3, 5, 4, 0, 0, 0, 0, 0,0,0,0 },
{ 3, 0, 0, 0, 3, 6, 0, 0, 0,0,0,0 },
{ 5, 0, 0, 2, 0, 0, 4, 0, 0,0,0,0 },
{ 4, 0, 2, 0, 1, 0, 0, 5, 0,0,0,0 },
{ 0, 3, 0, 2, 0, 2, 0, 0, 4,0,0,0 },
{ 0, 6, 0, 0, 2, 0, 0, 0, 0,5,0,0 },
{ 0, 0, 4, 0, 0, 0, 0, 3, 0,0,6,0 },
{ 0, 0, 0, 5, 0, 0, 3, 0, 6,0,7,0 },
{ 0, 0, 0, 0, 4, 0, 0, 6, 0,3,0,5 },
{ 0, 0, 0, 0, 0, 5, 0, 0, 3,0,0,9 },
{ 0, 0, 0, 0, 0, 0, 6, 7, 0,0,0,8 },
{ 0, 0, 0, 0, 0, 0, 0, 0, 5,9,8,0 }};

dijkstra(graph);

return 0;
}

I "C:\Users\AMIR KHAN\Desktop\c++\Dijkstra\main.exe" source vertex minimal distance a 0 a b 5 a a C d a a е f a a g a 5 7 9 9 9 12 15 a i j a a k execution time : 0.031 s Process returned 0 (0x0) Press any key to continue. 1 Type here to search 1 ))) 23:12 29-07-2020 E la ENG

source vertex minimal distance

a a 0
a b 3
a c 5
a d 4
a e 5
a f 7
a g 9
a h 9
a i 9
a j 12
a k 15