Multistage Graph (Shortest Path) - GeeksforGeeks (2023)

A Multistage graph is a directed, weighted 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).

The vertices of a multistage graph are divided into n number of disjoint subsets S = { S₁ , S_{2 ,} S₃_………..S_n }, where S₁ is the source and S_n is the sink ( destination ). The cardinality of S₁ and S_n are equal to 1. i.e., |S₁| = |S_n| = 1.
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 :-

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 given 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 targetas 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 andj 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.

We use top to bottom approach, and use dist[] array to store the value of overlapping sub-problem.

dist[i] will store the value of minimum distance from node i to node n-1 (target node).

Therefore, dist[0] will store minimum distance between from source node to target node.

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

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:

Time Complexity : O(n²)

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