Floyd warshall algorthim

// C Program for Floyd Warshall Algorithm 
#include<stdio.h> 
  
// Number of vertices in the graph 
#define V 4 
  
/* Define Infinite as a large enough value. This value will be used 
  for vertices not connected to each other */
#define INF 99999 
  
// A function to print the solution matrix 
void printSolution(int dist[][V]); 
  
// Solves the all-pairs shortest path problem using Floyd Warshall algorithm 
void floydWarshall (int graph[][V]) 
{ 
    /* dist[][] will be the output matrix that will finally have the shortest  
      distances between every pair of vertices */
    int dist[V][V], i, j, k; 
  
    /* Initialize the solution matrix same as input graph matrix. Or  
       we can say the initial values of shortest distances are based 
       on shortest paths considering no intermediate vertex. */
    for (i = 0; i < V; i++) 
        for (j = 0; j < V; j++) 
            dist[i][j] = graph[i][j]; 
  
    /* Add all vertices one by one to the set of intermediate vertices. 
      ---> Before start of an iteration, we have shortest distances between all 
      pairs of vertices such that the shortest distances consider only the 
      vertices in set {0, 1, 2, .. k-1} as intermediate vertices. 
      ----> After the end of an iteration, vertex no. k is added to the set of 
      intermediate vertices and the set becomes {0, 1, 2, .. k} */
    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); 
} 
  
/* A utility function to print solution */
void printSolution(int dist[][V]) 
{ 
    printf ("The 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"); 
    } 
} 
  
// driver program to test above function 
int main() 
{ 
   
    int graph[V][V] = { {0,   5,  INF, 10}, 
                        {INF, 0,   3, INF}, 
                        {INF, INF, 0,   1}, 
                        {INF, INF, INF, 0} 
                      }; 
  
    // Print the solution 
    floydWarshall(graph); 
    return 0; 
}