Floyd-warshall Algorithm

The Floyd-warshall algorithm is used to locate the shortest distance between each point. It needs to store edges using an adjacent matrix. This algorithm obtains the optimal path by considering the best sub-path.

Note that the path of an edge is not necessarily the best path.

• Start with any single-side path. The distance between all two points is the edge weight, or an infinite number. If there is no EDGE connection between the two points.
• For each pair of vertices u and v, check whether there is a vertex W which makes the path from u to W and then to V shorter than the known path. Update it.
• What is incredible is that you can get the result as long as you select the appropriate order.

// Dist (I, j) is the shortest path from node I to node J. For I do 1 to n do for J do 1 to n do dist (I, j) = weight (I, j) for k then 1 to n do // K is the "Media node" for I then 1 to n do for J then 1 to n do if (Dist (I, K) + dist (K, j) <dist (I, j) Then // is it a shorter path? Dist (I, j) = dist (I, K) + dist (K, J)

The efficiency of this algorithm is O (V3 ). It needs an adjacent matrix to store graphs.
This algorithm is easy to implement with just a few lines.

Even if the problem is to find the single-source shortest path, we recommend that you use this algorithm. If the time and space are allowed (as long as there is space in the bottom adjacent matrix, there is no problem in time ).

Sample Code:

# Include <stdio. h>
# Define NV 5
# Define Ma 9999999

Int main (void)
{
Int graph [NV] [NV] =
{
0, 1, 7, 2, 8,
1, 0, 3, 2, 2,
7, 3, 0, 4, 4,
2, 2, 4, 0, 3,
8, 2, 4, 3, 0
};
Int num = NV;
Int path [NV] [NV];
Int I, J, K;
For (I = 0; I <num; I ++ ){
For (j = 0; j <num; j ++)
Printf ("% d", graph [I] [J]);
Printf ("");
}
For (I = 0; I <num; I ++ ){
For (j = 0; j <num; j ++ ){
Path [I] [J] = graph [I] [J];
}
}

For (k = 0; k <num; k ++)
For (I = 0; I <num; I ++)
For (j = 0; j <num; j ++)
If (path [I] [J]> graph [I] [k] + graph [k] [J])
Path [I] [J] = graph [I] [k] + graph [k] [J];
For (I = 0; I <num; I ++)
{For (j = 0; j <num; j ++)
Printf ("shortest distance between <% d, % d>: % d", I + 1, J + 1, path [I] [J]);
}
Return 0;
}