Minimum cost path from source node to destination node via an intermediate node

Given an undirected weighted graph. The task is to find the minimum cost of the path from source node to the destination node via an intermediate node.

Note: If an edge is traveled twice, only once weight is calculated as cost.

Examples:

Input: source = 0, destination = 2, intermediate = 3;
Output: 6
The minimum cost path 0->1->3->1->2
The edge (1-3) occurs twice in the path, but its weight is
added only once to the answer.

Input: source = 0, destination = 2, intermediate = 1;
Output: 3
The minimum cost path is 0->1>2

Approach: Let suppose take a path P1 from Source to intermediate, and a path P2 from intermediate to destination. There can be some common edges among these 2 paths. Hence, the optimal path will always have the following form: for any node U, the walk consists of edges on the shortest path from Source to U, from intermediate to U, and from destination to U. Hence, if dist(a, b) is the cost of shortest path between node a and b, the required minimum cost path will be min{ dist(Source, U) + dist(intermediate, U) + dist(destination, U) } for all U. The Minimum distance of all nodes from Source, intermediate, and destination can be found by doing Dijkstra’s Shortest Path algorithm from these 3 nodes.

Below is the implementation of the above approach.

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// CPP program to find minimum distance between
// source and destination node and visiting
// of intermediate node is compulsory
#include <bits/stdc++.h>
using namespace std;
#define MAXN 100005
  
// to strore maped values of graph
vector<pair<int, int> > v[MAXN];
  
// to store distance of
// all nodes from the source node
int dist[MAXN];
  
// Dijkstra's algorithm to find
// shortest path from source to node
void dijkstra(int source, int n)
{
    // set all the vertices
    // distances as infinity
    for (int i = 0; i < n; i++)
        dist[i] = INT_MAX;
  
    // set all vertex as unvisited
    bool vis[n];
    memset(vis, false, sizeof vis);
  
    // make distance from source
    // vertex to source vertex is zero
    dist = 0;
  
    // // multiset do the job
    // as a min-priority queue
    multiset<pair<int, int> > s;
  
    // insert the source node with distance = 0
    s.insert({ 0, source });
  
    while (!s.empty()) {
        pair<int, int> p = *s.begin();
        // pop the vertex with the minimum distance
        s.erase(s.begin());
  
        int x = p.second;
        int wei = p.first;
  
        // check if the popped vertex
        // is visited before
        if (vis[x])
            continue;
  
        vis[x] = true;
  
        for (int i = 0; i < v[x].size(); i++) {
            int e = v[x][i].first;
            int w = v[x][i].second;
  
            // check if the next vertex
            // distance could be minimized
            if (dist[x] + w < dist[e]) {
  
                dist[e] = dist[x] + w;
  
                // insert the next vertex
                // with the updated distance
                s.insert({ dist[e], e });
            }
        }
    }
}
  
// function to add edges in graph
void add_edge(int s, int t, int weight)
{
    v[s].push_back({ t, weight });
    v[t].push_back({ s, weight });
}
  
// function to find the minimum shortest path
int solve(int source, int destination, 
               int intermediate, int n)
{
    int ans = INT_MAX;
  
    dijkstra(source, n);
  
    // store distance from source to
    // all other vertices
    int dsource[n];
    for (int i = 0; i < n; i++)
        dsource[i] = dist[i];
  
    dijkstra(destination, n);
    // store distance from destination
    // to all other vertices
    int ddestination[n];
    for (int i = 0; i < n; i++)
        ddestination[i] = dist[i];
  
    dijkstra(intermediate, n);
    // store distance from intermediate
    // to all other vertices
    int dintermediate[n];
    for (int i = 0; i < n; i++)
        dintermediate[i] = dist[i];
  
    // find required answer
    for (int i = 0; i < n; i++)
        ans = min(ans, dsource[i] + ddestination[i]
                                  + dintermediate[i]);
  
    return ans;
}
  
// Driver code
int main()
{
  
    int n = 4;
    int source = 0, destination = 2, intermediate = 3;
  
    // add edges in graph
    add_edge(0, 1, 1);
    add_edge(1, 2, 2);
    add_edge(1, 3, 3);
  
    // function call for minimum shortest path
    cout << solve(source, destination, intermediate, n);
  
    return 0;
}

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Output:

6

Time complexity: O((N + M) * logN), where N is number of nodes, M is number of edges.
Auxiliary Space: O(N+M)



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