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 thestopping conditionwill 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. #include<bits/stdc++.h> 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) continue; // 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] + dist[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; }

**Output:**

9

Time Complexity : O(n^{2})

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