All pair shortest path problem(Floyd Warshall Algorithm), with C Program Example

Floyd–Warshall algorithm is an algorithm for finding shortest paths in a weighted graph with positive or negative edge weights (but with no negative cycles)

Floyd Warshall Algorithm

We initialize the solution matrix same as the input graph matrix as a first step. Then we update the solution matrix by considering all vertices as an intermediate vertex. The idea is to one by one pick all vertices and update all shortest paths which include the picked vertex as an intermediate vertex in the shortest path. When we pick vertex number k as an intermediate vertex, we already have considered vertices {0, 1, 2, .. k-1} as intermediate vertices. For every pair (i, j) of source and destination vertices respectively, there are two possible cases.

• k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is.
• k is an intermediate vertex in shortest path from i to j. We update the value of dist[i][j] as dist[i][k] + dist[k][j].

Program in C :

#include<stdio.h>
int min(int,int);
void floyds(int p[10][10],int n)
{
 int i,j,k;
 for(k=1;k<=n;k++)
  for(i=1;i<=n;i++)
   for(j=1;j<=n;j++)
    if(i==j)
     p[i][j]=0;
    else
     p[i][j]=min(p[i][j],p[i][k]+p[k][j]);
}
int min(int a,int b)
{
 if(a<b)
  return(a);
 else
  return(b);
}
void main()
{
 int p[10][10],w,n,e,u,v,i,j;;
 printf("\n Enter the number of vertices:");
 scanf("%d",&n);
 printf("\n Enter the number of edges:\n");
 scanf("%d",&e);
 for(i=1;i<=n;i++)
 {
  for(j=1;j<=n;j++)
   p[i][j]=999;
 }
 for(i=1;i<=e;i++)
 {
  printf("\n Enter the end vertices of edge%d with its weight \n",i);
  scanf("%d%d%d",&u,&v,&w);
  p[u][v]=w;
 }
 printf("\n Matrix of input data:\n");
 for(i=1;i<=n;i++)
 {
  for(j=1;j<=n;j++)
   printf("%d \t",p[i][j]);
  printf("\n");
 }
 floyds(p,n);
 printf("\n Transitive closure:\n");
 for(i=1;i<=n;i++)
 {
  for(j=1;j<=n;j++)
   printf("%d \t",p[i][j]);
  printf("\n");
 }
 printf("\n The shortest paths are:\n");
 for(i=1;i<=n;i++)
  for(j=1;j<=n;j++)
  {
   if(i!=j)
    printf("\n <%d,%d>=%d",i,j,p[i][j]);
  }
}

• Output:

Please comment below in case of any problem found during running the code or any other doubts.