Shortest path from source to destination such that edge weights along path are alternatively increasing and decreasing

Given a connected graph with N vertices and M edges. The task is to find the shortest path from source to the destination vertex such that the difference between adjacent edge weights in the shortest path change from positive to negative and vice versa ( Weight(E1) > Weight(E2) < Weight(E3) …. ). If no such path exists then print -1. Examples:

Input: source = 4, destination = 3 Output: 19 4 – 2 – 1 – 3 (Edge Weights: 8, 3, 10) and 4 – 1 – 2 – 3 (Edge Weights: 6, 3, 10) are the only valid paths. Second path takes the minimum cost i.e. 19. Input: source = 2, destination = 4 Output: -1 No such path exists.

Approach: Here, We need to keep two copies of adjacent lists one for positive difference and other for negative difference. Take a Priority Queue as in Dijkstras Algorithm and keep four variables at a time i.e.,

1. cost: To store the cost of the path till current node.
2. stage: An integer variable to tell what element needs to be taken next, if the previous value was negative then a positive value needs to be taken else take negative.
3. weight: Weight of the last visited node.
4. vertex: Last visited vertex.

For every vertex push the adjacent vertices based on the required condition (value of stage). See the code for better understanding. Below is the implementation of the above approach: 

CPP

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 100005

// To store the graph
vector<pair<int, int> > incr[N], decr[N];
int _incr[N], _decr[N], shortest[N];

int n, m, src, dest, MAXI = 1LL << 30;

// Function to add edges
void Add_edge(int x, int y, int w)
{
    incr[x].push_back({ w, y });
    incr[y].push_back({ w, x });
    decr[x].push_back({ -w, y });
    decr[y].push_back({ -w, x });
}

// Function to find the shortest distance from
// source to destination
int Modified_Dijkstra()
{

    // Total cost, stage, weight of previous, vertex
    priority_queue<pair<pair<int, int>, pair<int, int> > > q;

    // Sort the edges
    for (int i = 1; i <= n; i++) {
        sort(incr[i].begin(), incr[i].end());
        sort(decr[i].begin(), decr[i].end());
    }

    for (int i = 1; i <= n; i++)
        shortest[i] = MAXI;

    // Push the source vertex
    q.push({ { 0, 0 }, { 0, src } });

    while (!q.empty()) {

        // Take the top element in the queue
        pair<pair<int, int>, pair<int, int> > FRONT = q.top();

        // Remove it from the queue
        q.pop();

        // Store all the values
        int cost = -FRONT.first.first;
        int stage = FRONT.first.second;
        int weight = FRONT.second.first;
        int v = FRONT.second.second;

        // Take the minimum cost for the vertex
        shortest[v] = min(shortest[v], cost);

        // If destination vertex has already been visited
        if (shortest[dest] != MAXI)
            break;

        // To make difference negative
        if (stage) {

            // Start from last not visited vertex
            for (int i = _incr[v]; i < incr[v].size(); i++) {

                // If we can take the ith vertex
                if (weight > incr[v][i].first)
                    q.push({ { -(cost + incr[v][i].first), 0 },
                             { incr[v][i].first, incr[v][i].second } });
                else {

                    // To keep the last not visited vertex
                    _incr[v] = i;
                    break;
                }
            }
        }

        // To make difference positive
        else {

            // Start from last not visited vertex
            for (int i = _decr[v]; i < decr[v].size(); i++) {

                // If we can take the ith vertex
                if (weight < -decr[v][i].first)
                    q.push({ { -(cost - decr[v][i].first), 1 },
                             { -decr[v][i].first, decr[v][i].second } });
                else {

                    // To keep the last not visited vertex
                    _decr[v] = i;
                    break;
                }
            }
        }
    }

    if (shortest[dest] == MAXI)
        return -1;

    return shortest[dest];
}

// Driver code
int main()
{
    n = 5, src = 4, dest = 3;

    // Adding edges
    Add_edge(4, 2, 8);
    Add_edge(1, 4, 6);
    Add_edge(2, 3, 10);
    Add_edge(3, 1, 10);
    Add_edge(1, 2, 3);
    Add_edge(3, 5, 3);

    cout << Modified_Dijkstra();

    return 0;
}

Output:

19

Time complexity:
The time complexity of the Modified Dijkstra’s algorithm is O((E+V)logV), where E is the number of edges and V is the number of vertices. In this implementation, we are using a priority queue to implement the algorithm, and sorting the edges takes O(ElogE) time. Therefore, the overall time complexity of this implementation is O((E+V)logV + ElogE).

Space complexity:
The space complexity of this implementation is O(E+V), where E is the number of edges and V is the number of vertices.