Warshall Algorithm - All pairs shortest path

//programme for floyd warshall algroithms -  all pair shortest path 
// Complexity - O(V ^ 3) where V is total vertices
#include<stdio.h>
#include<conio.h>

//number of vertices in ours graphh 
#define V 4 

/*
* Define Indefinite as large as possible to represent vertices
* not connected to eachother
*/

#define INF 99999

// a function to print solution matrix of  all pair shortest paths
void printSolution(int dist[][V]);

//solves the all- pair shortest path problem 
void floydWarshell(int graph[][V]){

//dist[][] will be the output matrix tht will  have shortest
//path 
int dist[V][V], i, j, k;

//initial solution matrix same as graph matrix
for (i = 0; i < V; i++)
for (j = 0; j < V; j++)
dist[i][j] = graph[i][j];

for (k = 0; k < V; k++)
{
// Pick all vertices as source one by one
for (i = 0; i < V; i++)
{
// Pick all vertices as destination for the
// above picked source
for (j = 0; j < V; j++)
{
// If vertex k is on the shortest path from
// i to j, then update the value of dist[i][j]
if (dist[i][k] + dist[k][j] < dist[i][j])
dist[i][j] = dist[i][k] + dist[k][j];
}
}
}

// Print the shortest distance matrix
printSolution(dist);








}


void printSolution(int dist[][V])
{
printf("Following matrix shows the shortest distances"
" between every pair of vertices \n");
for (int i = 0; i < V; i++)
{
for (int j = 0; j < V; j++)
{
if (dist[i][j] == INF)
printf("%7s", "INF");
else
printf("%7d", dist[i][j]);
}
printf("\n");
}
}



int main()
{
/* create the following weighted graph
     10
(0)------->(3)
|             /|\
   5   |              |1
|              | 
       \|/             |
      (1)------->(2)
     3         
 
*/
int graph[V][V] = 
{
{ 0, 5, INF, 10 },
{ INF, 0, 3, INF },
{ INF, INF, 0, 1 },
{ INF, INF, INF, 0 }
};

// Print the solution
floydWarshell(graph);
getchar();
return 0;
}