Jason
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Q:

floyd warshall algorithm c program

#include<stdio.h>#include<conio.h>int min(int,int);void floyds(int p[10][10],int n) {	int i,j,k;	for (k=1;k<=n;k++)	  for (i=1;i<=n;i++)	   for (j=1;j<=n;j++)	    if(i==j)	     p[i][j]=0; else	     p[i][j]=min(p[i][j],p[i][k]+p[k][j]);}int min(int a,int b) {	if(a<b)	  return(a); else	  return(b);}void main() {	int p[10][10],w,n,e,u,v,i,j;	;	clrscr();	printf("\n Enter the number of vertices:");	scanf("%d",&n);	printf("\n Enter the number of edges:\n");	scanf("%d",&e);	for (i=1;i<=n;i++) {		for (j=1;j<=n;j++)		   p[i][j]=999;	}	for (i=1;i<=e;i++) {		printf("\n Enter the end vertices of edge%d with its weight \n",i);		scanf("%d%d%d",&u,&v,&w);		p[u][v]=w;	}	printf("\n Matrix of input data:\n");	for (i=1;i<=n;i++) {		for (j=1;j<=n;j++)		   printf("%d \t",p[i][j]);		printf("\n");	}	floyds(p,n);	printf("\n Transitive closure:\n");	for (i=1;i<=n;i++) {		for (j=1;j<=n;j++)		   printf("%d \t",p[i][j]);		printf("\n");	}	printf("\n The shortest paths are:\n");	for (i=1;i<=n;i++)	  for (j=1;j<=n;j++) {		if(i!=j)		    printf("\n <%d,%d>=%d",i,j,p[i][j]);	}	getch();}
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