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

• Difficulty Level : Hard
• Last Updated : 24 Nov, 2021

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.

Examples:

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

Output: 21
Explanation:
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

Explanation:
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 need to be traversed and then traverse graph towards the leaf nodes with the maximum cost. And if the loop does not exist 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 a maximum cost leaf node and its children node(if it exists).
• check: boolean variable used as a flag to find a loop in the graph, its value changes to 0 when the 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 the adjacent node is found to be already visited and it is not the parent node then the loop is found and set the value of the check to 0.
4. Add the cost of all nodes of the loop to canTake.
5. After traversing adjacent nodes of the traversing node, no loop is found, then it represents the cost of the path leading from loop to leaf vertex and updates 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++

 // C++ program for the above approach #include using namespace std;   // To store the resulting // sum of the cost int canTake;   // To store largest // cost leaf vertex int best;   vector dp; vector vis;   // DFS Traversal to find the update // the maximum cost of from any // node to leaf int dfs(vector >& 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 >& 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;     dp.resize(n+1);       vis.resize(n+1);     // Cost Array     int cost[] = { 2, 2, 8, 6, 9 };       vector > g(n);       // Given Graph     g[0].push_back(1);     g[1].push_back(0);     g[0].push_back(2);     g[2].push_back(0);     g[0].push_back(3);     g[3].push_back(0);     g[1].push_back(2);     g[2].push_back(1);     g[1].push_back(4);     g[4].push_back(1);       // Given Source Node     int source = 1;       // Function Call     FindMaxCost(g, cost, source);     return 0; }

## Python3

 # Python3 program for the above approach N = 100000    # To store the resulting # sum of the cost canTake = 0    # To store largest # cost leaf vertex best = 0    dp = [0 for i in range(N)] vis = [0 for i in range(N)]    # DFS Traversal to find the update # the maximum cost of from any # node to leaf def dfs(g, cost, u, pre):           global canTake, best           # Mark vertex as visited     vis[u] = True        # Store vertex initial cost     dp[u] = cost[u]        # Initially assuming edge     # not to be traversed     check = 1        cur = cost[u]           for x in g[u]:            # Back edge found so,         # edge can be part of         # traversal         if (vis[x] and x != pre):             check = 0                       # New vertex is found         elif (not 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 (not 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 def FindMaxCost(g, cost, source):         # DFS Call     dfs(g, cost, source, -1)        # Print the maximum cost     print(canTake + best)       # Driver Code if __name__=='__main__':         n = 5     m = 5        # Cost Array     cost = [ 2, 2, 8, 6, 9 ]        g = [[] for i in range(n)]        # Given Graph     g[0].append(1)     g[1].append(0)     g[0].append(2)     g[2].append(0)     g[0].append(3)     g[3].append(0)     g[1].append(2)     g[2].append(1)     g[1].append(4)     g[4].append(1)        # Given Source Node     source = 1        # Function Call     FindMaxCost(g, cost, source)       # This code is contributed by rutvik_56

## Javascript



Output:

21

Time Complexity: O(N + M) where N is a number of vertices and M is the number of edges.
Auxiliary Space: O(N + M) where N is a number of vertices and M is a number of edges.

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