Floyd-warshall algorithm-Shortest path (suitable for node-dense graphs)

Since the time complexity of this algorithm is O (V3), in most cases it is not as good as the Dijkstra algorithm, the Dijkstra algorithm is suitable for the node evacuation diagram.

The example diagram is as follows:

Step 1 creates a table of shortest path results for nodes and edges ( directly accessible to the relationship), numeric representation of distance, INF representation unreachable

	1	2	3	4
1	0	8	INF	1
2	INF	0	1	INF
3 / td>	4	INF	0	INF
4	INF	2	9	0

Step2 Find All paths that pass 1, update the shortest path between two points

After 1 of the path is the combination of all the in and out paths, the total is in the X-out degree:

After 1 paths are:

First, 3-1-2.

At present min (3->2) is the INF, and Min (3->1->2) =4+8=12 so min (3->2) isA, because 2 is updated, so to recursively update all the shortest path from 3 to 2 points , The 2 point is only 3,min (3->3) is 0, so it does not need to be updated.

Article II, 3-1-4

At present min (3->4) is the INF, and Min (3->1->4) =4+1=5, so min (3->4) is 5, because 4 is updated, it is necessary to recursively update all the points from 3 to 4 of all the shortest path, and 4 of the points are 3 and 2:

Min (3->3) =0,min (3->2) =min (3->1->2) =12 > MIN (3->1->4->2) = 7, so MIN (3->2) =7

Find out all the results after 1 paths:

	1	2	3	4
1	0	8	Inf	1
2	Inf	0	1	Inf
3	4	7	0	5
4	Inf	2	9	0

Step3 Find all paths that go through 2.

1th article, 1->2->3

Because min (1->3) is INF, and Min (1->2->3) is 9, min (1->3) is 9, because 3 has an update, it is necessary to recursively update all the shortest paths from 1 to 3 points, because 3 can be up to a point of 1, While min (1->1) does not need to be updated, it is 0. Now look at the second path:

Article II, 4->2->3

So min (4->3) is 9, while min (4->2->3) is 3, so min (4->3) =3, because 3 has an update, need to recursively update the shortest path from 4 to 3 points, because 3 is up to 1, and min (4- >1) for Inf,min (4->2->3->1) is 7, so MIN (4->1) is 7. Since 1 has an update, continuation is recursive, 1 of the points are 4 and 2,min (4->4) remain 0; min (4->2) is 2, and min (4->2->3->1->2) =2+1+4+8=15 is greater than 2, so no update is required.

After finding all paths that pass 2, the result is:

	1	2	3	4
1	0	8	9	1
2	Inf	0	1	Inf
3	4	7	0	5
4	7	2	3	0

Step4 Find all paths that go through 3.

1th article, 4->3->1

Min (4->1) is 7, and min (4->3->1) is 13, so there is no need to update

Article II, 2->3->1

Because min (2->1) is INF, and Min (2->3->1) is 5, you need to update min (2->1) to 5. Since the update is 1, it is necessary to update all paths from 2 to 1 points, 1 of the points to 4 and 2,min (2->2) do not need to be updated; min (2->4) is INF, and Min (2->3->1->4) is 6, so min (2->4) is 6. Because the update is 4, it is necessary to recursively update the path from 2 to 4 points, 4 to 2 and 3,min (2->2) to 0;min (2->3) to 1 less than MIN (2->3->1->4->3) =1+4+1+9= 15, so there is no need to update.

So after this step, the result table is:

	1	2	3	4
1	0	8	9	1
2	5	0	1	6
3	4	7	0	5
4	7	2	3	0

Finally, find the path that passes through 4.

First, 1->4->2.

Min (1->2) is 8, and min (1->4->2) is 3, so min (1->2) is updated to 3, due to the update of 2, it is necessary to update all the shortest path from 1 to 2 points, and 2 to the point of 3,min (1-> 3) currently 9, and Min (1->4->2->3) is 4, so min (1->3) is updated to 4, due to the update of 3, so recursively update all the shortest path from 1 to 3 points, 3 of the point is 1, and min (1->1 ) does not require an update to remain at 0.

Article II, 1->4->3

Min (1->3) is 4, and Min (1->4->3) is 10, so no update is required. So the end result is:

	1	2	3	4
1	0	3	4	1
2	5	0	1	6
3	4	7	0	5
4	7	2	3	0

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Floyd-warshall algorithm-Shortest path (suitable for node-dense graphs)