Given a network of cities and the distances between them,the task is to find the shortest path from a city(source) to all other cities connected to it.The network of cities with their distances is represented as a weighted digraph.It is also known as single-source,shortest path problem.

Algorithm-

N is the number of vertices labeled {0,1,2,3….N-1} of the weighted digraph.cost[0:N-1][0:N-1] is the cost matrix of the graph.If there is no edge from i to j,then cost[i][j]=INT_MAX.If i equals j,then cost[i][j]=0.

Vertex 0 is the source.

V is the set of N cities

T={0}; //T is initialized by adding the source vertex

//distance[] is initialized to the cost of the edges connecting vertex i with the source vertex 0.

for(i=1 to N)

{

distance[i]=cost(0,i);

}

for(i=0 to N-2)

{

Choose a vertex u in V-T such that distance[u] is a minimum;

Add u to T;

for each vertex w in V-T

{

distance[w]=minimum(distance[w],distance[u]+cost[u,w]);

}

}

Its time complexity is O(N^2) where N is the number of cities.

**Code-**

[cpp]

for(i=0;i<n;i++)

{

b[i]=0;

distance[i]=cost[0][i];

}

//b[i]=1 if i is in set T and b[i]=0 if it is in set V-T.

b[0]=1;//T is initialized

z=1;//counter initialized to 1

while(z!=n)

{

min=INT_MAX;

for(i=0;i<n;i++)

{

if((b[i]==0)&&(distance[i]<min))

{

min=distance[i];

u=i;

}

}

b[u]=1;

z++;

for(w=0;w<n;w++)

{

if((cost[u][w]!=0) && (b[w]==0) && (distance[u]!=INT_MAX) && (distance[u]+cost[u][w]<distance[w]))

{

distance[w]=distance[u]+cost[u][w];

}

}

}

//printing the shortest distances

for(i=0;i<n;i++)

printf("%d ",distance[i]);

[/cpp]

Example-

If the cost matrix is-

@ denotes INT_MAX

0 20 @ 40 110

@ 0 60 @ @

@ @ 0 @ 20

@ @ 30 0 70

@ @ @ @ 0

The output is-

distance[0]=0

distance[1]=20

distance[2]=70

distance[3]=40

distance[4]=90