Multistage Graph (Shortest Path)

A Multistage graph is a directed graph in which the nodes can be divided into a set of stages such that all edges are from a stage to next stage only (In other words there is no edge between vertices of same stage and from a vertex of current stage to previous stage).

We are give a multistage graph, a source and a destination, we need to find shortest path from source to destination. By convention, we consider source at stage 1 and destination as last stage.

Following is an example graph we will consider in this article :-

Now there are various strategies we can apply :-

  • The Brute force method of finding all possible paths between Source and Destination and then finding the minimum. That’s the WORST possible strategy.
  • Dijkstra’s Algorithm of Single Source shortest paths. This method will find shortest paths from source to all other nodes which is not required in this case. So it will take a lot of time and it doesn’t even use the SPECIAL feature that this MULTI-STAGE graph has.
  • Simple Greedy Method – At each node, choose the shortest outgoing path. If we apply this approach to the example graph give above we get the solution as 1 + 4 + 18 = 23. But a quick look at the graph will show much shorter paths available than 23. So the greedy method fails !
  • The best option is Dynamic Programming. So we need to find Optimal Sub-structure, Recursive Equations and Overlapping Sub-problems.

Optimal Substructure and Recursive Equation :-

We define the notation :- M(x, y) as the minimum cost to T(target node) from Stage x, Node y.

Shortest distance from stage 1, node 0 to 
destination, i.e., 7 is M(1, 0).

// From 0, we can go to 1 or 2 or 3 to
// reach 7.              
M(1, 0) = min(1 + M(2, 1),
              2 + M(2, 2),
              5 + M(2, 3))

This means that our problem of 0 —> 7 is now sub-divided into 3 sub-problems :-

So if we have total 'n' stages and target
as T, then the stopping condition  will be :-
M(n-1, i) = i ---> T + M(n, T) = i ---> T

Recursion Tree and Overlapping Sub-Problems:-
So, the hierarchy of M(x, y) evaluations will look something like this :-

In M(i, j), i is stage number and
j is node number

                   M(1, 0)
           /          |         \                             
          /           |          \                            
       M(2, 1)      M(2, 2)        M(2, 3)
    /      \        /     \         /    \
M(3, 4)  M(3, 5)  M(3, 4)  M(3, 5) M(3, 6)  M(3, 6)
 .         .       .       .          .        .
 .         .       .       .          .        .
 .         .       .       .          .        .

So, here we have drawn a very small part of the Recursion Tree and we can already see Overlapping Sub-Problems. We can largely reduce the number of M(x, y) evaluations using Dynamic Programming.

Implementation details:
The below implementation assumes that nodes are numbered from 0 to N-1 from first stage (source) to last stage (destination). We also assume that the input graph is multistage.

// CPP program to find shortest distance
// in a multistage graph.
using namespace std;
#define N 8
#define INF INT_MAX
// Returns shortest distance from 0 to
// N-1.
int shortestDist(int graph[N][N]) {
    // dist[i] is going to store shortest
    // distance from node i to node N-1.
    int dist[N];
    dist[N-1] = 0;
    // Calculating shortest path for
    // rest of the nodes
    for (int i = N-2 ; i >= 0 ; i--)
        // Initialize distance from i to
        // destination (N-1)
        dist[i] = INF;
        // Check all nodes of next stages
        // to find shortest distance from
        // i to N-1.
        for (int j = i ; j < N ; j++)
            // Reject if no edge exists
            if (graph[i][j] == INF)
            // We apply recursive equation to
            // distance to target through j.
            // and compare with minimum distance 
            // so far.
            dist[i] = min(dist[i], graph[i][j] +
    return dist[0];
// Driver code
int main()
    // Graph stored in the form of an
    // adjacency Matrix
    int graph[N][N] =
      {{INF, 1, 2, 5, INF, INF, INF, INF},
       {INF, INF, INF, INF, 4, 11, INF, INF},
       {INF, INF, INF, INF, 9, 5, 16, INF},
       {INF, INF, INF, INF, INF, INF, 2, INF},
       {INF, INF, INF, INF, INF, INF, INF, 18},
       {INF, INF, INF, INF, INF, INF, INF, 13},
       {INF, INF, INF, INF, INF, INF, INF, 2}};
    cout << shortestDist(graph);
    return 0;



Time Complexity : O(n2)

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