// C++ Program for Floyd Warshall Algorithm
#include <bits/stdc++.h>
using namespace std;
// 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])
{
cout<<"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)
cout<<"INF"<<" ";
else
cout<<dist[i][j]<<" ";
}
cout<<endl;
}
}
// Driver code
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;
}
// This code is contributed by rathbhupendra
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