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

? Because this algorithm has a time complexity of O (V3). In most cases it is not as good as the Dijkstra algorithm. The Dijkstra algorithm is suitable for node evacuation diagrams.

The demo sample diagram is as follows:

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Step 1 Creates a table of the shortest path results for nodes and edges ( directly up to the relationship ). The value represents the distance. INF indicates unreachable

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

Step2 find all paths that go through 1. Update the shortest path between two points

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

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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, it is necessary to recursively update all the shortest path from 3 to 2 points . The 2 point is only 3,min (3->3) is 0, so there is no need to update.

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 from 3 to 4 of all the shortest path of the point, 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 the results of all 1 paths:

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?	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.

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1th article, 1->2->3

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

The second article. 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 of the point is 1, and MIN (4->1) is Inf,min (4->2->3->1) 7. So MIN (4->1) is 7. Since 1 has an update, the continuation is recursive, 1 of the points are 4 and 2,min (4->4) remain 0, the min (4->2) is 2, and min (4->2->3->1->2) =2+1+4+8=15 is greater than 2. So there is no need to update.

After you find all the paths that go through 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. Therefore , the need to update MIN (2->1) is 5.

Due to the update of 1, it is necessary to update all the paths from 2 to 1 points, 1 points to 4 and 2,min (2->2) do not need to be updated; now MIN (2->4) is INF. and Min (2->3->1->4) is 6. So MIN (2->4) is 6. Due to the update of 4, it is necessary to recursively update the paths 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. It does not need to be updated.

So after this step, the result table is:

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?	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.

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First article. 1->4->2

Min (1->2) is 8, while min (1->4->2) is 3, so min (1->2) is updated to 3, because 2 is updated. It is necessary to update the shortest path from 1 to 2 points, and 2 to 3. MIN (1->3) is now 9. Min (1->4->2->3) is 4, so min (1->3) is updated to 4. Due to the update of 3. Therefore recursive update all from 1 to 3 the shortest path of the point, 3 of the point is 1, and min (1->1) does not need to update to remain 0.

Article II, 1->4->3

Min (1->3) is 4, while min (1->4->3) is 10. So there is no need to update.

So finally the 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)