Maximum cost path from source node to destination node via at most K intermediate nodes

Given a directed weighted graph consisting of N vertices and an array Edges[][], with each row representing two vertices connected by an edge and the weight of that edge, the task is to find the path with the maximum sum of weights from a given source vertex src to a given destination vertex dst, made up of at most K intermediate vertices. If no such path exists, then print -1.

Examples:

Input: N = 3, Edges[][] = {{0, 1, 100}, {1, 2, 100}, {0, 2, 500}}, src = 0, dst = 2, K = 0
Output: 500
Explanation:

Path 0 → 2: The path with maximum weight and at most 0 intermediate nodes is of weight 500.

Input: N = 3, Edges[][] = {{0, 1, 100}, {1, 2, 100}, {0, 2, 500}}, src = 0, dst = 2, K = 0
Output: 500
Explanation:

Path 0 → 2: The path with maximum weight and at most 1 intermediate node is of weight 500.

Approach: The given problem can be solved by using BFS(Breadth-First Search) Traversal. Follow the steps below to solve the problem:

• Initialize the variable, say ans, to store the maximum distance between the source and the destination node having at most K intermediates nodes.
• Initialize an adjacency list of the graph using the edges.
• Initialize an empty queue and push the source vertex into it. Initialize a variable, say lvl, to store the number of nodes present in between src and dst.
• While the queue is not empty and lvl is less than K + 2 perform the following steps:
  • Store the size of the queue in a variable, say S.
  • Iterate over the range [1, S] and perform the following steps:
    • Pop the front element of the queue and store it in a variable, say T.
    • If T is the dst vertex, then update the value of ans as the maximum of ans and the current distance T.second.
    • Traverse through all the neighbors of the current popped node and check if the distance of its neighbor is greater than the current distance or not. If found to be true, then push it in the queue and update its distance.
  • Increase the value of lvl by 1.
• After completing the above steps, print the value of ans as the resultant maximum distance.

Below is the implementation of the above approach:

C++

// C++ program for the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to find the longest distance
// from source to destination with at
// most K intermediate nodes
int findShortestPath(
    int n, vector<vector<int> >& edges,
    int src, int dst, int K)
{
    // Initialize the adjacency list
    vector<vector<pair<int, int> > > adjlist(
        n, vector<pair<int, int> >());

    // Initialize a queue to perform BFS
    queue<pair<int, int> > q;

    unordered_map<int, int> mp;

    // Store the maximum distance of
    // every node from source vertex
    int ans = INT_MIN;

    // Initialize adjacency list
    for (int i = 0; i < edges.size(); i++) {

        auto edge = edges[i];

        adjlist[edge[0]].push_back(
            make_pair(edge[1], edge[2]));
    }

    // Push the first element into queue
    q.push({ src, 0 });

    int level = 0;

    // Iterate until the queue becomes empty
    // and the number of nodes between src
    // and dst vertex is at most to K
    while (!q.empty() && level < K + 2) {

        // Current size of the queue
        int sz = q.size();

        for (int i = 0; i < sz; i++) {

            // Extract the front
            // element of the queue
            auto pr = q.front();

            // Pop the front element
            // of the queue
            q.pop();

            // If the dst vertex is reached
            if (pr.first == dst)
                ans = max(ans, pr.second);

            // Traverse the adjacent nodes
            for (auto pr2 : adjlist[pr.first]) {

                // If the distance is greater
                // than the current distance
                if (mp.find(pr2.first)
                        == mp.end()
                    || mp[pr2.first]
                           > pr.second
                                 + pr2.second) {

                    // Push it into the queue
                    q.push({ pr2.first,
                             pr.second
                                 + pr2.second });
                    mp[pr2.first] = pr.second
                                    + pr2.second;
                }
            }
        }

        // Increment the level by 1
        level++;
    }

    // Finally, return the maximum distance
    return ans != INT_MIN ? ans : -1;
}

// Driver Code
int main()
{
    int n = 3, src = 0, dst = 2, k = 1;
    vector<vector<int> > edges
        = { { 0, 1, 100 },
            { 1, 2, 100 },
            { 0, 2, 500 } };

    cout << findShortestPath(n, edges,
                             src, dst, k);

    return 0;
}

Python3

# Python3 program for the above approach
from collections import deque

# Function to find the longest distance
# from source to destination with at
# most K intermediate nodes
def findShortestPath(n, edges, src, dst, K):

    # Initialize the adjacency list
    adjlist = [[] for i in range(n)]

    # Initialize a queue to perform BFS
    q = deque()

    mp = {}

    # Store the maximum distance of
    # every node from source vertex
    ans = -10**9

    # Initialize adjacency list
    for i in range(len(edges)):
        edge = edges[i]
        adjlist[edge[0]].append([edge[1],
                                 edge[2]])

    # Push the first element into queue
    q.append([src, 0])

    level = 0

    # Iterate until the queue becomes empty
    # and the number of nodes between src
    # and dst vertex is at most to K
    while (len(q) > 0 and level < K + 2):

        # Current size of the queue
        sz = len(q)

        for i in range(sz):

            # Extract the front
            # element of the queue
            pr = q.popleft()

            # Pop the front element
            # of the queue
            # q.pop()

            # If the dst vertex is reached
            if (pr[0] == dst):
                ans = max(ans, pr[1])

            # Traverse the adjacent nodes
            for pr2 in adjlist[pr[0]]:

                # If the distance is greater
                # than the current distance
                if ((pr2[0] not in mp) or
                  mp[pr2[0]] > pr[1] + pr2[1]):

                    # Push it into the queue
                    q.append([pr2[0], pr[1] + pr2[1]])
                    mp[pr2[0]] = pr[1] + pr2[1]

        # Increment the level by 1
        level += 1

    # Finally, return the maximum distance
    return ans if ans != -10**9 else -1

# Driver Code
if __name__ == '__main__':

    n, src, dst, k = 3, 0, 2, 1

    edges= [ [ 0, 1, 100 ],
             [ 1, 2, 100 ],
             [ 0, 2, 500 ] ]

    print(findShortestPath(n, edges,src, dst, k))

# This code is contributed by mohit kumar 29

Output:

500

Time Complexity: O(N + E)
Auxiliary Space: O(N)