Skip to content

Instantly share code, notes, and snippets.

@zSANSANz
Last active April 5, 2021 18:47
Show Gist options
  • Save zSANSANz/071012052e531dcaaec9b6f74b87d1c4 to your computer and use it in GitHub Desktop.
Save zSANSANz/071012052e531dcaaec9b6f74b87d1c4 to your computer and use it in GitHub Desktop.
Floyd Warshall Algorithm on C++
// 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 a 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 a 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 ("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()
{
/* Let us 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
floydWarshall(graph);
return 0;
}
@aminPial
Copy link

aminPial commented Mar 6, 2020

dude, are you trying to just show off someone that you know all of those algorithm by just Copy Pasting from other sources! XD

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment