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 = 21Input: 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:
- Perform DFS traversal with flag variable check set to ‘1’ initially denoting no loop found.
- Simultaneously build the dp[] for each node with the maximum cost updated till that traversed node.
- 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.
- Add the cost of all nodes of the loop to canTake.
- 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.
- 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 <bits/stdc++.h>using namespace std;// To store the resulting// sum of the costint canTake;// To store largest// cost leaf vertexint best;vector<int> dp;vector<bool> vis;// DFS Traversal to find the update// the maximum cost of from any// node to leafint 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 twiceint FindMaxCost(vector<vector<int> >& g, int* cost, int source){ // DFS Call dfs(g, cost, source, -1); // Print the maximum cost cout << canTake + best;}// Driver Codeint main(){ int n = 5, m = 5; dp.resize(n+1); vis.resize(n+1); // Cost Array int cost[] = { 2, 2, 8, 6, 9 }; vector<vector<int> > 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;} |
Java
// Java program for the above approachimport java.util.*;class GFG{ static int N = 100000;// To store the resulting// sum of the coststatic int canTake;// To store largest// cost leaf vertexstatic int best;static int []dp = new int[N];static boolean []vis = new boolean[N];// DFS Traversal to find the update// the maximum cost of from any// node to leafstatic boolean dfs(Vector<Integer> []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 boolean check = true; int cur = cost[u]; for(int x : g[u]) { // Back edge found so, // edge can be part of // traversal if (vis[x] && x != pre) { check = false; } // 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) ? false : true; // Adds parent and its // children cost cur = Math.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 = Math.max(best, dp[u]); } return check;}// Function to find the maximum cost// from source vertex such that no// two edges is traversed twicestatic void FindMaxCost(Vector<Integer> [] g, int []cost, int source){ // DFS call dfs(g, cost, source, -1); // Print the maximum cost System.out.print(canTake + best);}// Driver Codepublic static void main(String[] args){ int n = 5, m = 5; // Cost Array int cost[] = { 2, 2, 8, 6, 9 }; @SuppressWarnings("unchecked") Vector<Integer> []g = new Vector[n]; for(int i = 0; i < g.length; i++) g[i] = new Vector<Integer>(); // Given Graph g[0].add(1); g[1].add(0); g[0].add(2); g[2].add(0); g[0].add(3); g[3].add(0); g[1].add(2); g[2].add(1); g[1].add(4); g[4].add(1); // Given Source Node int source = 1; // Function call FindMaxCost(g, cost, source);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 program for the above approachN = 100000 # To store the resulting# sum of the costcanTake = 0 # To store largest# cost leaf vertexbest = 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 leafdef 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 twicedef FindMaxCost(g, cost, source): # DFS Call dfs(g, cost, source, -1) # Print the maximum cost print(canTake + best) # Driver Codeif __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 |
C#
// C# program for // the above approachusing System;using System.Collections.Generic;class GFG{ static int N = 100000;// To store the resulting// sum of the coststatic int canTake;// To store largest// cost leaf vertexstatic int best;static int []dp = new int[N];static bool []vis = new bool[N];// DFS Traversal to find the update// the maximum cost of from any// node to leafstatic bool dfs(List<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 = true; int cur = cost[u]; foreach(int x in g[u]) { // Back edge found so, // edge can be part of // traversal if (vis[x] && x != pre) { check = false; } // 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) ? false : true; // Adds parent and its // children cost cur = Math.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 = Math.Max(best, dp[u]); } return check;}// Function to find the maximum cost// from source vertex such that no// two edges is traversed twicestatic void FindMaxCost(List<int> [] g, int []cost, int source){ // DFS call dfs(g, cost, source, -1); // Print the maximum cost Console.Write(canTake + best);}// Driver Codepublic static void Main(String[] args){ int n = 5, m = 5; // Cost Array int []cost = {2, 2, 8, 6, 9}; List<int> []g = new List<int>[n]; for(int i = 0; i < g.Length; i++) g[i] = new List<int>(); // Given Graph g[0].Add(1); g[1].Add(0); g[0].Add(2); g[2].Add(0); g[0].Add(3); g[3].Add(0); g[1].Add(2); g[2].Add(1); g[1].Add(4); g[4].Add(1); // Given Source Node int source = 1; // Function call FindMaxCost(g, cost, source);}}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript program for // the above approach var N = 100000;// To store the resulting// sum of the costvar canTake = 0;// To store largest// cost leaf vertexvar best = 0;var dp = Array(N).fill(0);var vis = Array(N).fill(false);// DFS Traversal to find the update// the maximum cost of from any// node to leaffunction dfs(g, cost, u, pre){ // Mark vertex as visited vis[u] = true; // Store vertex initial cost dp[u] = cost[u]; // Initially assuming edge // not to be traversed var check = true; var cur = cost[u]; for(var x of g[u]) { // Back edge found so, // edge can be part of // traversal if (vis[x] && x != pre) { check = false; } // 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) ? false : true; // Adds parent and its // children cost cur = Math.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) { // push cost of vertex // to the answer canTake += cost[u]; } else { // Updates the largest // cost leaf vertex best = Math.max(best, dp[u]); } return check;}// Function to find the maximum cost// from source vertex such that no// two edges is traversed twicefunction FindMaxCost(g, cost, source){ // DFS call dfs(g, cost, source, -1); // Print the maximum cost document.write(canTake + best);}// Driver Codevar n = 5, m = 5;// Cost Arrayvar cost = [2, 2, 8, 6, 9];var g = Array.from(Array(n), ()=>Array());// Given Graphg[0].push(1);g[1].push(0);g[0].push(2);g[2].push(0);g[0].push(3);g[3].push(0);g[1].push(2);g[2].push(1);g[1].push(4);g[4].push(1);// Given Source Nodevar source = 1;// Function callFindMaxCost(g, cost, source);</script> |
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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