#include<stdio.h>

#include<assert.h>

/* maxVertices represents maximum number of vertices that can be present in the graph. */

#define maxVertices 100

#define INF 123456789

/* Input Format: Graph is directed and weighted. First two integers must be number of vertces and edges

which must be followed by pairs of vertices which has an edge between them.

*/

int min(int a,int b)

{

return (a<b)?a:b;

}

void init(int distance[maxVertices][maxVertices])

{

int iter,jter;

for(iter=0;iter<maxVertices;iter++)

{

for(jter=0;jter<maxVertices;jter++)

{

if(iter==jter)

{

distance[iter][jter] = 0;

}

else

{

distance[iter][jter] = INF;

}

}

}

}

void FloydWarshall(int distance[maxVertices][maxVertices],int vertices)

{

int from,to,via;

for(from=0;from<vertices;from++)

{

for(to=0;to<vertices;to++)

{

for(via=0;via<vertices;via++)

{

distance[from][to] = min(distance[from][to],

distance[from][via]+distance[via][to]);

}

}

}

}

int main()

{

int graph[maxVertices][maxVertices],size[maxVertices]={0},visited[maxVertices]={0};

int distance[maxVertices][maxVertices];

int vertices,edges,iter,jter;

/* vertices represent number of vertices and edges represent number of edges in the graph. */

scanf("%d%d",&vertices,&edges);

/*initialize distance between all pairs as infinity*/

init(distance);

int vertex1,vertex2,weight;

/* Here graph[i][j] represent the weight of edge joining i and j */

for(iter=0;iter<edges;iter++)

{

scanf("%d%d%d",&vertex1,&vertex2,&weight);

assert(vertex1>=0 && vertex1<vertices);

assert(vertex2>=0 && vertex2<vertices);

graph[vertex1][vertex2] = weight;

distance[vertex1][vertex2] = weight;

}

FloydWarshall(distance,vertices);

for(iter=0;iter<vertices;iter++)

{

for(jter=0;jter<vertices;jter++)

{

if(distance[iter][jter]>=INF)

{

printf("The shortest distance between %d and %d is Infinity\n",iter,jter);

}

else

{

printf("The shortest distance between %d and %d is %d\n",iter,jter,distance[iter][jter]);

}

}

}

return 0;

}

**Rough Notes about the Algorithm as implemented in the code above:**
**Input Format: **Graph is directed and weighted. First two integers must be number of vertices and
edges which must be followed by pairs of vertices which has an edge between them.
**maxVertices** represents maximum number of vertices that can be present in the graph.
**vertices** represent number of vertices and edges represent number of edges in the graph.
**graph[i][j]** represent the weight of edge joining i and j.
**size[maxVertices] **is initialed to{0}, represents the size of every vertex i.e. the number of
**edges** corresponding to the vertex.
**visited[maxVertices]={0}** represents the vertex that have been visited.
**distance[maxVertices][maxVertices]** represents the weight of the edge between the two vertices or distance between two vertices.
Initialize the distance between two vertices using **init()** function.

**init() function**- It takes the distance matrix as an argument.
For iter=0 to maxVertices – 1
For jter=0 to maxVertices – 1
if(iter == jter)
distance[iter][jter] = 0 //Distance between two same vertices is 0
else
distance[iter][jter] = INF//Distance between different vertices is INF
jter + 1
iter + 1
Where, INF is a very large integer value.
Initialize and input the graph.

**Call FloydWarshall function.**
• It takes the distance matrix (distance[maxVertices][maxVertices]) and number of
vertices as argument (vertices).
• Initialize integer type from, to, via
For from=0 to vertices-1
For to=0 to vertices-1
For via=0 to vertices-1
distance[from][to] = min(distance[from][to],distance[from][via]+distance[via][to])
via + 1
to + 1
from + 1

This finds the minimum distance from from vertex to to vertex using the min
function. It checks it there are intermediate vertices between the from and to
vertex that form the shortest path between them
• min function returns the minimum of the two integers it takes as argument.
Output the distance between every two vertices.