Floyd algorithm, one of the shortest path algorithms

Floyd algorithm

The Floyd algorithm can be used to solve the shortest path problem between any two vertices.

Core formula:

　　　　　　Edge [I] [J] = min {edge [I] [J], edge [I] [k] + edge [k] [J]}.

That is to say, we can relax by inserting the vertex between I and j and comparing the path size.

First, we define a two-dimensional array edge [maxn] [maxn] to store Graph Information.

 

After the edge array of the graph is initialized

 

It is equivalent to the distance between any two points that cannot pass through other points.

Code1:

1 // After vertex 2 for (I = 1; I <= N; I ++) 3 for (j = 1; j <= N; j ++) 4 If (E [I] [J]> E [I] [1] + E [1] [J]) E [I] [J] = E [I] [1] + E [1] [J];

In this example, vertex 1 is allowed as the center point to relax the distance and save the result after relaxation.

Code2:

1 // After vertex 2 for (I = 1; I <= N; I ++) 3 for (j = 1; j <= N; j ++) 4 If (E [I] [J]> E [I] [2] + E [2] [J]) E [I] [J] = E [I] [2] + E [2] [J];

Allow vertex 1 and vertex 2 to relax and save as intermediate points. (Not always relaxed !)

.....

Core code of Floyd:

 

1 for(k=1;k<=n;k++)2     for(i=1;i<=n;i++)3         for(j=1;j<=n;j++)4             if(e[i][j]>e[i][k]+e[k][j])5                  e[i][j]=e[i][k]+e[k][j];

 

The basic idea of this Code is: In the beginning, it is allowed to transit only through vertex 1, and then only through vertex 1 and vertex 2 ...... 1 ~ All vertices on N are transitioned to find the shortest path between any two points. In a word, the shortest distance from vertex I to vertex J is only the first K.

Time Complexity: O (N ^ 3)

Some of the image text is taken from the blog of Aha lei.