Floyd algorithm of shortest circuit

1. Introduction

Floyd algorithm only five lines of code, the code is simple, three for loop can solve the problem, so its time complexity is O (n^3), can find multi-source shortest path problem.

2. Thought:

The basic idea of the Floyd algorithm is as follows: from any node A to any node B the shortest path is 2 possible, 1 is directly from a to b,2 is from a through a number of nodes X to B. Therefore, we assume that dis (AB) is the shortest path distance from Node A to Node B, for each node x we check whether dis (AX) + dis (XB) < DIS (AB) is established, and if so, prove that the path from a to x and then to B is shorter than the path of a directly to B, We set up dis (AB) = Dis (AX) + dis (XB) so that when we go through all the nodes X,dis (AB), the distance from the shortest path of A to B is recorded.

As an example: known,

If now only allowed through the 1th vertex, the shortest distance between any two points, just Judge E[i][1]+e[1][j] is smaller than e[i][j]. E[I][J] represents the distance from the I vertex to the J number vertex. E[I][1]+E[1][J] Represents the sum of the distances from vertex i to number 1th, and from vertex number 1th to vertex J. Where I is the 1~n loop, and J is also the 1~n loop, the code is implemented as follows.

 for (i=1; i<=n; i++) {    for (j=1; j<=n; j + +)    {        if  (E[i][j] > e[i][1]+e[1][j])             = e[i][1]+e[1][j];}    }

Next, continue to ask for the shortest distance between any two points that only allow two vertices of 1 and 2nd. In the result of allowing only the shortest distance of any two points through the 1th vertex, it is also possible to determine whether the distance between vertex I and the J vertex can be shortened if the number 2nd vertex is passed. That is to judge E[i][2]+e[2][j] is smaller than e[i][j], the code is implemented as follows.

//after vertex number 1th for(i=1; i<=n; i++)     for(j=1; j<=n; J + +)        if(E[i][j] > e[i][1]+e[1][j]) e[i][j]=e[i][1]+e[1][j];//after vertex number 2nd for(i=1; i<=n; i++)     for(j=1; j<=n; J + +)        if(E[i][j] > e[i][2]+e[2][j]) e[i][j]=e[i][2]+e[2][J];

Finally, all vertices are allowed to be brokered, and the code is as follows:

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

The basic idea of this code is that it is only allowed to go through the number 1th vertices at first, and then only the 1 and 2nd vertices are allowed to relay ... The shortest distance between any two points is allowed through all vertices of the 1~n number. Same as above

3. Code templates:

#include <stdio.h>#defineINF 0x3f3f3f3fintmap[ +][ +];intMain () {intk,i,j,n,m; //reads N and m,n to indicate the number of vertices, m represents the number of edgesscanf"%d%d",&n,&m); //Initialize     for(i=1; i<=n; i++)         for(j=1; j<=n; J + +)            if(i==j) Map[i][j]=0; ElseMap[i][j]=inf; intA,b,c; //Read in Edge     for(i=1; i<=m; i++) {scanf (" %d%d%d",&a,&b,&c); MAP[A][B]=c;//This is a graph with a direction    }    //Floyd-warshall algorithm Core statement     for(k=1; k<=n; k++)         for(i=1; i<=n; i++)             for(j=1; j<=n; J + +)                if(map[i][j]>map[i][k]+Map[k][j]) map[i][j]=map[i][k]+Map[k][j]; //outputs The final result, the final two-dimensional array, even if the shortest distance between two points     for(i=1; i<=n; i++)    {         for(j=1; j<=n; J + +) {printf ("%10d", Map[i][j]); } printf ("\ n"); }    return 0;}

Floyd algorithm of shortest circuit