C++ Program to Find SSSP (Single Source Shortest Path) in DAG (Directed Acyclic Graphs)

C++Server Side ProgrammingProgramming

This is a C++ program to find SSSP (Single Source Shortest Path) in DAG (Directed Acyclic Graphs) using Dijkstra Algorithm to find out from the first node in graph to every other node with the shortest path length showed beside each pair of vertices.

Algorithm

Begin
   Take the elements of the graph as input.
   function shortestpath():
   Initialize the variables
   a[i] = 1
   d[i] = 0
   s[i].from = 0
   Initialize a loop for i = 0 to 3 do
      if b[0][i] == 0
         continue
      else
         d[i] = b[0][i]
         s[i].from = 0
      done
   done
   Initialize a loop while (c < 4)
   initialize min = INFINITY
   for i = 0 to 3 do
      if min <= d[i] or d[i] == 0 or a[i] == 1
         continue
      else if min > d[i]
         min = d[i]
   done
   for loop int k = 0 to 3 do
      if (min == d[k])
         t = k
         break
      else
         continue
      done
      Initialize a[t] = 1
      for j = 0 to 3
         if a[j] == 1 or b[t][j] == 0
            continue
         else if a[j] != 1
            if d[j] > (d[t] + b[t][j])
               d[j] = d[t] + b[t][j]
               s[i].from = t
      done
      Increment c
   done
   For loop i = 0 to 3
      Print minimum cost from node1 to node2.
   done
End

Example

#include <iostream>
using namespace std;
#define INFINITY 9999
struct node {
   int from;
} s[4];
int c = 0;
void djikstras(int *a, int b[][4], int *d) {
   int i = 0, j, min, t;
   a[i] = 1;
   d[i] = 0;
   s[i].from = 0;
   for (i = 0; i < 4;i++) {
      if (b[0][i] == 0) {
         continue;
      } else {
         d[i] = b[0][i];
         s[i].from = 0;
      }
   }
   while (c < 4) {
      min = INFINITY;
      for (i = 0; i < 4; i++) {
         if (min <= d[i] || d[i] == 0 || a[i] == 1) {
            continue;
         } else if (min > d[i]) {
            min = d[i];
         }
      }
      for (int k = 0; k < 4; k++) {
         if (min == d[k]) {
            t = k;
            break;
         } else {
            continue;
         }
      }
      a[t] = 1;
      for (j = 0; j < 4; j++) {
         if (a[j] == 1 || b[t][j] == 0) {
            continue;
         } else if (a[j] != 1) {
            if (d[j] > (d[t] + b[t][j])) {
               d[j] = d[t] + b[t][j];
               s[i].from = t;
            }
         }
      }
      c++;
   }
   for (int i = 0; i < 4; i++) {
      cout<<"from node "<<s[i].from<<" cost is:"<<d[i]<<endl;
   }
}
int main() {
   int a[4];
   int d[4];
   for(int k = 0; k < 4; k++) {
      d[k] = INFINITY;
   }
   for (int i = 0; i < 4; i++) {
      a[i] = 0;
   }
   int b[4][4];
   for (int i = 0;i < 4;i++) {
      cout<<"enter values for "<<(i+1)<<" row"<<endl;
      for(int j = 0;j < 4;j++) {
         cin>>b[i][j];
      }
   }
   djikstras(a,b,d);
}

Output

enter values for 1 row
0
1
3
2
enter values for 2 row
2
1
3
0
enter values for 3 row
2
3
0
1
enter values for 4 row
1
3
2
0
from node 0 cost is:0
from node 0 cost is:1
from node 0 cost is:3
from node 0 cost is:2
raja
Published on 26-Apr-2019 08:10:50
Advertisements