An Example:

```Input:
graph[][] = { {0,   5,  INF, 10},
{INF,  0,  3,  INF},
{INF, INF, 0,   1},
{INF, INF, INF, 0} }
which represents the following graph
10
(0)------->(3)
|         /|\
5 |          |
|          | 1
\|/         |
(1)------->(2)
3
Note that the value of graph[i][j] is 0 if i is equal to j
And graph[i][j] is INF (infinite) if there is no edge from vertex i to j.

Output:
Shortest distance matrix
0      5      8      9
INF      0      3      4
INF    INF      0      1
INF    INF    INF      0
```

Floyd Warshall Algorithm :

We initialize the solution matrix same as the input graph matrix as a first step. Then we are updating the solution matrix by considering all vertices as an intermediate vertex. The thought is to one by one pick all vertices and update all shortest paths which include the selected 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 each pair (i, j) of source and destination vertices respectively, there are two possible cases.
1) k is not an intermediate vertex in shortest path from i to j. We keep the value of dist[i][j] as it is.
2) 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].

The following digram is taken from the Cormen book. It shows the above optimal substructure property in the all-pairs shortest path problem. C implementation of the Floyd Warshall algorithm is given as follows :

 // Program for Floyd Warshall Algorithm#include// Number of vertices in the graph#define V 4/* Define Infinite as a large enough value. This value will be usedfor vertices not connected to each other */#define INF 99999// A function to print the solution matrixvoid printSolution(int dist[][V]);// Solves the all-pairs shortest path problem using Floyd Warshall algorithmvoid floydWarshell (int graph[][V]){/* dist[][] will be the output matrix that will finally have the shortestdistances between every pair of vertices */int dist[V][V], i, j, k;/* Initialize the solution matrix same as input graph matrix. Orwe can say the initial values of shortest distances are basedon 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 allpairs of vertices such that the shortest distances consider only thevertices in set {0, 1, 2, .. k-1} as intermediate vertices.—-> After the end of a iteration, vertex no. k is added to the set ofintermediate vertices and the set becomes {0, 1, 2, .. k} */for (k = 0; k < V; k++){// Pick all vertices as source one by onefor (i = 0; i < V; i++){// Pick all vertices as destination for the// above picked sourcefor (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 matrixprintSolution(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”);elseprintf (“%7d”, dist[i][j]);}printf(“\n”);}}// driver program to test above functionint main(){/* Let us create the following weighted graph10(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 solutionfloydWarshell(graph);return 0;}

Output:

```Following matrix shows the shortest distances between every pair of vertices
0      5      8      9
INF      0      3      4
INF    INF      0      1
INF    INF    INF      0
```

Time Complexity: O(V^3)

The given program only displays the shortest distances. We can modify the solution to display the shortest paths also by storing the predecessor info in a separate 2D matrix.
Also, the value of INF can be taken as INT_MAX from limits.h to make sure that we handle maximum possible value. When we take INF as INT_MAX, we need to change the if condition in the given code to avoid arithmetic overflow.