Johnson’s Algorithm in C++

This is a C++ Program to implement Johnson’s Algorithm. Johnson’s algorithm helps to find shortest path between given source and destination nodes.

Here is source code of the C++ Program to Implement Johnson’s Algorithm. The C++ program is successfully compiled and run on a Linux system. The program output is also shown below.

```
#include<iostream>
```
```
#include<conio.h>
```
```
 
```
```
using namespace std;
```
```
 
```
```
int min(int a, int b);
```

int cost[10][10], a[10][10], i, j, k, c;

```
 
```
```
int min(int a, int b)
```
```
{
```
```
    if (a < b)
```
```
        return a;
```
```
    else
```
```
        return b;
```
```
}
```
```
 
```
```
int main(int argc, char **argv)
```
```
{
```
```
    int n, m;
```
```
    cout << "Enter no of vertices";
```
```
    cin >> n;
```
```
    cout << "Enter no of edges";
```
```
    cin >> m;
```
```
    cout << "Enter the\nEDGE Cost\n";
```
```
    for (k = 1; k <= m; k++)
```
```
    {
```
```
        cin >> i >> j >> c;
```
```
        a[i][j] = cost[i][j] = c;
```
```
    }
```
```
    for (i = 1; i <= n; i++)
```
```
        for (j = 1; j <= n; j++)
```
```
        {
```
```
            if (a[i][j] == 0 && i != j)
```
```
                a[i][j] = 31999;
```
```
        }
```
```
    for (k = 1; k <= n; k++)
```
```
        for (i = 1; i <= n; i++)
```
```
            for (j = 1; j <= n; j++)
```

                a[i][j] = min(a[i][j], a[i][k] + a[k][j]);

```
    cout << "Resultant adj matrix\n";
```
```
    for (i = 1; i <= n; i++)
```
```
    {
```
```
        for (j = 1; j <= n; j++)
```
```
        {
```
```
            if (a[i][j] != 31999)
```
```
                cout << a[i][j] << " ";
```
```
        }
```
```
        cout << "\n";
```
```
    }
```
```
    return 0;
```
```
}
```

Output:

$ g++ JohnsonsShortestPath.cpp
$ a.out
 
Enter no of vertices 3
Enter no of edges 5
Enter the
EDGE Cost
1 2 4
2 1 6
1 3 11
3 1 3
2 3 2
Resultant adj matrix
0 4 6 
5 0 2 
3 7 0 
 
------------------
(program exited with code: 0)
Press return to continue

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