a). Design and implement C/C++ Program to solve All-Pairs Shortest Paths problem using Floyd's algorithm
#include<stdio.h>
#define INF 999
int min(int a,int b)
{
return(a<b)?a:b;
}
void floyd(int p[][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++)
p[i][j]=min(p[i][j],p[i][k]+p[k][j]);
}
void main()
{
int a[10][10],n,i,j;
printf("\nEnter the n value:");
scanf("%d",&n);
printf("\nEnter the graph data:\n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&a[i][j]);
floyd(a,n);
printf("\nShortest path matrix\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%d ",a[i][j]);
printf("\n");
}
}
OUTPUT:
Enter the n value:4
Enter the graph data:
0 999 3 999
2 0 999 999
999 7 0 1
6 999 999 0
Shortest path matrix
0 10 3 4
2 0 5 6
7 7 0 1
6 16 9 0
b) Design and implement C/C++ Program to find the transitive closure using Warshal's algorithm
#include<stdio.h>
void warsh(int p[][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++)
p[i][j]=p[i][j] || p[i][k] && p[k][j];
}
int main()
{
int a[10][10],n,i,j;
printf("\nEnter the n value:");
scanf("%d",&n);
printf("\nEnter the graph data:\n");
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
scanf("%d",&a[i][j]);
warsh(a,n);
printf("\nResultant path matrix\n");
for(i=1;i<=n;i++)
{
for(j=1;j<=n;j++)
printf("%d ",a[i][j]);
printf("\n");
}
return 0;
}
OUTPUT:
Enter the n value:4
Enter the graph data:
0 1 0 0
0 0 0 1
0 0 0 0
1 0 1 0
Resultant path matrix
1 1 1 1
1 1 1 1
0 0 0 0
1 1 1 1