__Environment:__ C++, any version

### Introduction

Graph algorithms work on nodes and edges. Nodes and edges are represented in many ways. This well-known algorithm uses a connectivity matrix. The matrix is a square matrix with number of columns and rows same as many as there are Nodes. The diagonal elements are set to zero. The offdiagonals are 1 if connected 0 if not connected. (You could also use distances if you are dealing with something like road distances.) This matrix is modified to make the cost of the disconnected nodes to infinity. You can make slight modifications so that the inputs take STL vector< vector<int> > as a matrix data structure. You could drop the lower half of the matrix; you can also consider using the upper half for the original data and using the lower half for the distance matrix and save entirely on an extra copy of the matrix.

Since the earlier version, I have added code that stores the nodes in the path in addition to the length, so that the actual paths are also computed in addition to the path lengths, thus doing what Djikstra’s algorithm was primarily meant to do, as well.

The output of the algorithm is a matrix of the shortest distances between every pair of points and a comma-separated list of nodes representing the paths.

– L. Shyamal

Bangalore, India

### The Code

#include <stdio.h>

#include <stdlib.h>

#include <string>

using namespace std;// Floyd’s All pairs shortest path algorithm (O (n^3) )

// input is adjacency matrix output is matrix of shortest paths

// C is the adjacency matrix

// n is the order of the square matrix

// A is the all pairs shortest paths matrix

// we assume that A is allocated by the caller

int ComputeFloydAPSP(int *C, int n, int *A, string* s)

{int i,j,k;

// set all connected positions to 1

// and all unconnected positions to infinity

for (i=0; i<n; i++)

{

for (j=0; j<n; j++)

{

if ( *(C+i*n+j) == 0)

{

*(A+i*n+j) = 999999999; // Does this look like infinity?

}

else

{

*(A+i*n+j) = 1;

char sz[3]=””;

sprintf(sz,”%d”,j+1);

s[i*n+j]=sz;

}

}

}// set the diagonals to zero

for (i=0; i<n; i++)

{

*(A+i*n+i) = 0;

}// for each route via k from i to j pick any better routes and

// replace A[i][j] path with sum of paths i-k and j-k

for (k=0; k<n; k++)

{

for (i=0; i<n; i++)

{

for (j=0; j<n; j++)

{

if ( *(A+i*n+k) + *(A+k*n+j) < *(A+i*n+j) )

{

// A[i][j] = A[i][k] + A[k][j];

*(A+i*n+j) = *(A+i*n+k)+ *(A+k*n+j);

//s[i*n+j]=s[i*n+j]+s[i*n+k]+s[k*n+j];

s[i*n+j]=s[i*n+k]+”,”+s[k*n+j];

}

}

}

}return 0;

} // Floyd’s algorithm// this is for testing Floyd’s algorithm

// demonstrates the allocation and

// deallocation of memory for the matrices

void FloydTest()

{// allocate the entire matrix in one linear array

// trying to allocate it as an array of pointers to arrays

// didn’t quite work, possibly because the [] and * notation

// aren’t quite replaceable

int n=10;

int *C=new int[n*n];C[0 ]=0;C[1 ]=1;C[2 ]=0;C[3 ]=0;C[4 ]=0;C[5 ]=0;C[6 ]=0;C[7 ]=0;

C[8 ]=0;C[9 ]=0;

C[10]=1;C[11]=0;C[12]=1;C[13]=0;C[14]=0;C[15]=0;C[16]=1;C[17]=0;

C[18]=0;C[19]=0;

C[20]=0;C[21]=1;C[22]=0;C[23]=1;C[24]=0;C[25]=0;C[26]=0;C[27]=0;

C[28]=0;C[29]=1;

C[30]=0;C[31]=0;C[32]=1;C[33]=0;C[34]=1;C[35]=0;C[36]=0;C[37]=0;

C[38]=0;C[39]=0;

C[40]=0;C[41]=0;C[42]=0;C[43]=1;C[44]=0;C[45]=1;C[46]=0;C[47]=0;

C[48]=0;C[49]=0;

C[50]=0;C[51]=0;C[52]=0;C[53]=0;C[54]=1;C[55]=0;C[56]=1;C[57]=0;

C[58]=0;C[59]=0;

C[60]=0;C[61]=1;C[62]=0;C[63]=0;C[64]=0;C[65]=1;C[66]=0;C[67]=1;

C[68]=0;C[69]=0;

C[70]=0;C[71]=0;C[72]=0;C[73]=0;C[74]=0;C[75]=0;C[76]=1;C[77]=0;

C[78]=1;C[79]=1;

C[80]=0;C[81]=0;C[82]=0;C[83]=0;C[84]=0;C[85]=0;C[86]=0;C[87]=1;

C[88]=0;C[89]=0;

C[90]=0;C[91]=0;C[92]=1;C[93]=0;C[94]=0;C[95]=0;C[96]=0;C[97]=1;

C[98]=0;C[99]=0;int* A = new int[n*n];

string* s = new string[n*n];

printf(“Initial matrixn”);

for(int i=0;i<n;i++)

{

for(int j=0;j<n;j++)

{

printf(“%d “,*(C+i*n+j));

}

printf(“n”);

}ComputeFloydAPSP (C,n,A,s);

printf(“Final shortest distancesn”);

for(i=0;i<n;i++)

{

for(int j=0;j<n;j++)

{

printf(“%d “,*(A+i*n+j));

}

printf(“n”);

}

printf(“End of All pairs Shortest pathsn”);

for(i=0;i<n;i++)

{

for(int j=i+1;j<n;j++)

{

printf(“path from %d to %d is %sn”,i+1,j+1,s[i*n+j].c_str());

}

printf(“n”);

}delete [] A;

delete [] C;

delete [] s;} // end of FloydTest

void main()

{

FloydTest();

char c;

scanf(“%c”,&c);

}