# 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 given 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.

## C++

`// 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},` ` ` `{INF, INF, INF, INF, INF, INF, INF, INF}};` ` ` `cout << shortestDist(graph);` ` ` `return` `0;` `}` |

## Java

`// Java program to find shortest distance` `// in a multistage graph.` `class` `GFG` `{` ` ` `static` `int` `N = ` `8` `;` ` ` `static` `int` `INF = Integer.MAX_VALUE;` ` ` `// Returns shortest distance from 0 to` ` ` `// N-1.` ` ` `public` `static` `int` `shortestDist(` `int` `[][] graph)` ` ` `{` ` ` `// dist[i] is going to store shortest` ` ` `// distance from node i to node N-1.` ` ` `int` `[] dist = ` `new` `int` `[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] = Math.min(dist[i], graph[i][j]` ` ` `+ dist[j]);` ` ` `}` ` ` `}` ` ` `return` `dist[` `0` `];` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String[] args)` ` ` `{` ` ` `// Graph stored in the form of an` ` ` `// adjacency Matrix` ` ` `int` `[][] graph = ` `new` `int` `[][]{{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` `}};` ` ` `System.out.println(shortestDist(graph));` ` ` `}` `}` `// This code has been contributed by 29AjayKumar` |

## Python3

`# Python3 program to find shortest` `# distance in a multistage graph.` `# Returns shortest distance from` `# 0 to N-1.` `def` `shortestDist(graph):` ` ` `global` `INF` ` ` `# dist[i] is going to store shortest` ` ` `# distance from node i to node N-1.` ` ` `dist ` `=` `[` `0` `] ` `*` `N` ` ` `dist[N ` `-` `1` `] ` `=` `0` ` ` `# Calculating shortest path` ` ` `# for rest of the nodes` ` ` `for` `i ` `in` `range` `(N ` `-` `2` `, ` `-` `1` `, ` `-` `1` `):` ` ` `# 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` `j ` `in` `range` `(N):` ` ` ` ` `# 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` `N ` `=` `8` `INF ` `=` `999999999999` `# Graph stored in the form of an` `# adjacency Matrix` `graph ` `=` `[[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` `]]` `print` `(shortestDist(graph))` `# This code is contributed by PranchalK` |

## C#

`// C# program to find shortest distance` `// in a multistage graph.` `using` `System;` ` ` `class` `GFG` `{` ` ` `static` `int` `N = 8;` ` ` `static` `int` `INF = ` `int` `.MaxValue;` ` ` ` ` `// Returns shortest distance from 0 to` ` ` `// N-1.` ` ` `public` `static` `int` `shortestDist(` `int` `[,] graph) {` ` ` ` ` `// dist[i] is going to store shortest` ` ` `// distance from node i to node N-1.` ` ` `int` `[] dist = ` `new` `int` `[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] = Math.Min(dist[i], graph[i,j] +` ` ` `dist[j]);` ` ` `}` ` ` `}` ` ` ` ` `return` `dist[0];` ` ` `}` ` ` ` ` `// Driver code` ` ` `static` `void` `Main()` ` ` `{` ` ` `// Graph stored in the form of an` ` ` `// adjacency Matrix` ` ` `int` `[,] graph = ` `new` `int` `[,]` ` ` `{{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}};` ` ` ` ` `Console.Write(shortestDist(graph));` ` ` `}` `}` `// This code is contributed by DrRoot_` |

## Javascript

`<script>` `// JavaScript program to find shortest distance` `// in a multistage graph.` `let N = 8;` `let INF = Number.MAX_VALUE;` `// Returns shortest distance from 0 to` ` ` `// N-1.` `function` `shortestDist(graph)` `{` ` ` `// dist[i] is going to store shortest` ` ` `// distance from node i to node N-1.` ` ` `let dist = ` `new` `Array(N);` ` ` ` ` `dist[N - 1] = 0;` ` ` ` ` `// Calculating shortest path for` ` ` `// rest of the nodes` ` ` `for` `(let 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` `(let 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] = Math.min(dist[i], graph[i][j]` ` ` `+ dist[j]);` ` ` `}` ` ` `}` ` ` ` ` `return` `dist[0];` `}` `let graph = [[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]];` `document.write(shortestDist(graph));` `// This code is contributed by rag2127` `</script>` |

**Output:**

9

**Time Complexity :** O(n^{2})

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