Maximum cost path in an Undirected Graph such that no edge is visited twice in a row

Given an undirected graph having N vertices and M edges and each vertex is associated with a cost and a source vertex S is given. The task is to find the maximum cost path from source vertex S such that no edge is visited consecutively 2 or more times.


Input: N = 5, M = 5, source = 1, cost[] = {2, 2, 8, 6, 9}, Below is the given graph:

Output: 21
The maximum cost path matrix is given as:
1 -> 2 -> 0 -> 1 -> 4
Cost = 2 + 8 + 2 + 2 + 9 = 21

Input: N = 8, M = 8, source = 3, cost[] = {10, 11, 4, 12, 3, 4, 7, 9}

Output: 46
The maximum cost path matrix is given as:
3 -> 0 -> 2 -> 1 -> 7

Approach: The idea is to check if there exists a loop exists in the graph, then all vertices of the loop needs to be traversed and then traverse graph towards the leaf nodes with maximum cost. And if loop does not exists then then the problem statement converts to find maximum cost path in any directed graph.
Below are the declaration used in the program:

  • dp[i]: stores the total cost to traverse the node ‘i’ and all it’s children node.
  • vis[i]: marks the nodes which have been visited.
  • canTake: stores the resultant sum of all node of maximum cost path excluding the leaf vertex and its children node, if it exists.
  • best: stores the cost of maximum cost leaf node and its children node(if it exists).
  • check: boolean variable used as a flag to find loop in the graph, its value changes to 0 when loop is found.

Below are the steps:

  1. Perform DFS traversal with flag variable check set to ‘1’ initially denoting no loop found.
  2. Simultaneously build the dp[] for each node with the maximum cost updated till that traversed node.
  3. If adjacent node is found to be already visited and it is not parent node then loop is found and set value of check to 0.
  4. Add cost of all nodes of the loop to canTake.
  5. After traversing adjacent nodes of the traversing node, no loop is found, then it represent cost of path leading from loop to leaf vertex and update best to dp[i] if dp[i] is greater than best.
  6. After traversal of the graph, print the sum of canTake and best.

Below is the implementation of the above approach:






// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
// To store the resulting
// sum of the cost
int canTake;
// To store largest
// cost leaf vertex
int best;
int dp[N];
bool vis[N];
// DFS Traversal to find the update
// the maximum cost of from any
// node to leaf
int dfs(vector<vector<int> >& g,
        int* cost, int u, int pre)
    // Mark vertex as visited
    vis[u] = true;
    // Store vertex initial cost
    dp[u] = cost[u];
    // Initially assuming edge
    // not to be traversed
    bool check = 1;
    int cur = cost[u];
    for (auto& x : g[u]) {
        // Back edge found so,
        // edge can be part of
        // traversal
        if (vis[x] && x != pre) {
            check = 0;
        // New vertex is found
        else if (!vis[x]) {
            // Bitwise AND the current
            // check with the returned
            // check by the previous
            // DFS Call
            check &= dfs(g, cost, x, u);
            // Adds parent and its
            // children cost
            cur = max(cur,
                      cost[u] + dp[x]);
    // Updates total cost of parent
    // including child nodes
    dp[u] = cur;
    // Edge is part of the cycle
    if (!check) {
        // Add cost of vertex
        // to the answer
        canTake += cost[u];
    else {
        // Updates the largest
        // cost leaf vertex
        best = max(best, dp[u]);
    return check;
// Function to find the maximum cost
// from source vertex such that no
// two edges is traversed twice
int FindMaxCost(vector<vector<int> >& g,
                int* cost, int source)
    // DFS Call
    dfs(g, cost, source, -1);
    // Print the maximum cost
    cout << canTake + best;
// Driver Code
int main()
    int n = 5, m = 5;
    // Cost Array
    int cost[] = { 2, 2, 8, 6, 9 };
    vector<vector<int> > g(n);
    // Given Graph
    // Given Source Node
    int source = 1;
    // Function Call
    FindMaxCost(g, cost, source);
    return 0;




Time Complexity: O(N + M) where N is number of vertices and M is number of edges
Auxillary Space: O(N + M) where N is number of vertices and M is number of edges

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