Maximum sum possible for every node by including it in a segment of N-Ary Tree
Given an N-Ary tree containing N nodes and an array weight [ ] that denotes the weight of the nodes which can be positive or negative, the task for every node is to print the maximum sum possible by a sequence of nodes including the current node.
Examples:
Input: N = 7
weight[] = [-8, 9, 7, -4, 5, -10, -6]
N-Ary tree:
-8
/ \
9 7
/ \ /
-4 5 -10
/
-6
Output: 13 14 13 10 14 3 4
Explanations:
Node -8: [-8 + 9 + 7 + 5] = 13
Node 9: [9 + 5] = 14
Node 3: [7 + (-8) + 9 + 5] = 13
Node 4: [-4 + 9 + 5] = 10
Node: [5 + 9] = 14
Node 6: [-10 + 7 + (-8) + 9 + 5] = 3
Node 7: [-6 + (-4) + 9 + 5] = 4
Input: N = 6
weight[] = [2, -7, -5, 8, 4, -10]
N-Ary tree:
2
/ \
-7 -5
/ \ \
8 4 -10
Output: 7 7 2 8 7 -8Approach: This problem can be solved using Dp on Trees technique by applying two DFS.
- Apply the first DFS to store the maximum sum possible for every node by including them in a sequence with their respective successors. Store the maximum possible sum in dp1[]. array.
- Maximum possible value for each node in the first DFS can be obtained by:
dp1[node] += maximum(0, dp1[child1], dp1[child2], …)
- Perform the second Dfs to update the maximum sum for each node in dp1[] by including them in a sequence with their ancestors also. The maximum values stored in dp2[] for every node are the required answers.
- Maximum possible value for each node in the second DFS can be obtained by:
dp2[node] = dp1[node] + maximum(0, maxSumAncestors)
Best answer can be obtained by including or excluding the maximum sum possible for its ancestors
where maxSumAncestors = dp2[parent] – maximum(0, dp1[node]), i.e. including or excluding contribution of the maximum sum of the current node stored in dp1[]
Refer to the pictorial explanation for better understanding:
Below is the implementation of the above approach:
C++
// C++ program to calculate the maximum// sum possible for every node by including// it in a segment of the N-Ary Tree#include <bits/stdc++.h>using namespace std;// Stores the maximum// sum possible for every node// by including them in a segment// with their successorsint dp1[100005];// Stores the maximum// sum possible for every node// by including them in a segment// with their ancestorsint dp2[100005];// Store the maximum sum// for every node by// including it in a// segment with its successorsvoid dfs1(int u, int par, vector<int> g[], int weight[]){ dp1[u] = weight[u]; for (auto c: g[u]) { if (c != par) { dfs1(c, u, g, weight); dp1[u] += max(0, dp1); } }}// Update the maximum sums// for each node by including// them in a sequence with// their ancestorsvoid dfs2(int u, int par, vector<int> g[], int weight[]){ // Condition to check, // if current node is not root if (par != 0) { int maxSumAncestors = dp2[par] - max(0, dp1[u]); dp2[u] = dp1[u] + max(0, maxSumAncestors); } for (auto c: g[u]) { if (c != par) { dfs2(c, u, g, weight); } }}// Add edgesvoid addEdge(int u, int v, vector<int> g[]){ g[u].push_back(v); g[v].push_back(u);}// Function to find the maximum// answer for each nodevoid maxSumSegments(vector<int> g[], int weight[], int n){ // Compute the maximum sums // with successors dfs1(1, 0, g, weight); // Store the computed maximums for (int i = 1; i <= n; i++) { dp2[i] = dp1[i]; } // Update the maximum sums // by including their // ancestors dfs2(1, 0, g, weight);}// Print the desired resultvoid printAns(int n){ for (int i = 1; i <= n; i++) { cout << dp2[i] << " "; }}// Driver Programint main(){ // Number of nodes int n = 6; int u, v; // graph vector<int> g[100005]; // Add edges addEdge(1, 2, g); addEdge(1, 3, g); addEdge(2, 4, g); addEdge(2, 5, g); addEdge(3, 6, g); addEdge(4, 7, g); // Weight of each node int weight[n + 1]; weight[1] = -8; weight[2] = 9; weight[3] = 7; weight[4] = -4; weight[5] = 5; weight[6] = -10; weight[7] = -6; // Compute the max sum // of segments for each // node maxSumSegments(g, weight, n); // Print the answer // for every node printAns(n); return 0;} |
Python3
# Python3 program to calculate the maximum# sum possible for every node by including# it in a segment of the N-Ary Tree # Stores the maximum# sum possible for every node# by including them in a segment# with their successorsdp1 = [0 for i in range(100005)] # Stores the maximum sum possible# for every node by including them# in a segment with their ancestorsdp2 = [0 for i in range(100005)] # Store the maximum sum for every# node by including it in a# segment with its successorsdef dfs1(u, par, g, weight): dp1[u] = weight[u] for c in g[u]: if (c != par): dfs1(c, u, g, weight) dp1[u] += max(0, dp1) # Update the maximum sums# for each node by including# them in a sequence with# their ancestorsdef dfs2(u, par, g, weight): # Condition to check, # if current node is not root if (par != 0): maxSumAncestors = dp2[par] - max(0, dp1[u]) dp2[u] = dp1[u] + max(0, maxSumAncestors) for c in g[u]: if (c != par): dfs2(c, u, g, weight) # Add edgesdef addEdge(u, v, g): g[u].append(v) g[v].append(u)# Function to find the maximum# answer for each nodedef maxSumSegments(g, weight, n): # Compute the maximum sums # with successors dfs1(1, 0, g, weight) # Store the computed maximums for i in range(1, n + 1): dp2[i] = dp1[i] # Update the maximum sums # by including their # ancestors dfs2(1, 0, g, weight)# Print the desired resultdef printAns(n): for i in range(1, n): print(dp2[i], end = ' ') # Driver codeif __name__=='__main__': # Number of nodes n = 7 u = 0 v = 0 # Graph g = [[] for i in range(100005)] # Add edges addEdge(1, 2, g) addEdge(1, 3, g) addEdge(2, 4, g) addEdge(2, 5, g) addEdge(3, 6, g) addEdge(4, 7, g) # Weight of each node weight=[0 for i in range(n + 1)] weight[1] = -8 weight[2] = 9 weight[3] = 7 weight[4] = -4 weight[5] = 5 weight[6] = -10 weight[7] = -6 # Compute the max sum # of segments for each # node maxSumSegments(g, weight, n) # Print the answer # for every node printAns(n)# This code is contributed by pratham76 |
Output:
13 14 13 10 14 3
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