Sum of the distances from every node to all other nodes is maximum
Given a tree with N nodes and N-1 edges with root at 1 and given an array of N-1 integers. The task is to assign weights to the edges in the tree such that the sum of the distances from every node to all other nodes is maximum.
Examples:
Input:
Output: 46
Assign the edge 1-2 with weight 5
Assign the edge 2-3 with weight 7
Assign the edge 3-4 with weight 1
The distance of node 1 from the nodes 2, 3, 4 is {5, 5+7, 5+7+1}
The distance of node 2 from the nodes 3, 4 is {7, 7+1}
The distance of node 3 from the node 4 is {1}Input:
Output: 94
Approach: The problem can be solved using Combinations, DFS, DP on trees and Greedy logic. Since we need to assign weights to edges in the tree, hence assigning the maximum weight to the edge which occurs the maximum number of times in all the paths will be the way to get the maximum sum. In order to find the number of times an edge occurs in all the paths possible, we need to know the number of nodes on both the side of the edge. Let c1 and c2 be the count of the number of nodes on the left and right side, then the number of times the edge occurs in all paths will be c1 * c2. Sort all the possible values of c1 * c2 in ascending order. Assign the maximum weight to the maximum c1 * c2 value, and to the others in the same way. We can follow the below steps to get the number of nodes on the left side and on the right side of an edge:
- Run a dfs starting from the root, and initialize a dp[] array which stores the count of the nodes in the subtree of a given node.
- Iterate for every possible edge, and find the number of nodes in the both the side of the edges.
- To find the number of nodes in both the sides, find out the smaller value of dp[node1] or dp[node2], where node1 and node2 are the nodes on the either side of the edge
- If one side has min(dp[node1], dp[node2]), then the other side will have (N – min(dp[node1], dp[node2])).
Below is the implementation of the above approach:
C++
// C++ program to implement the // above approach #include <bits/stdc++.h> using namespace std; // Function to add an edge to the tree void addEdge(vector<pair< int , int > >& edges, list< int >* tree, int x, int y) { edges.push_back({ x, y }); tree[x].push_back(y); tree[y].push_back(x); } // Function to run DFS and calculate the // height of the subtree below it void dfs(vector<pair< int , int > >& edges, list< int >* tree, int node, int parent, int dp[]) { // Initially initialize with 1 dp[node] = 1; // Traverse for all nodes connected to node for ( auto it : tree[node]) { // If node is not parent // then recall dfs if (it != parent) { dfs(edges, tree, it, node, dp); // Add the size of the // subtree beneath it dp[node] += dp[it]; } } } // Function to assign weights to edges // to maximize the final sum int maximizeSum( int a[], vector<pair< int , int > >& edges, list< int >* tree, int n) { // Initialize it which stores the // height of the subtree beneath it int dp[n + 1] = { 0 }; // Call the DFS function to dfs(edges, tree, 1, 0, dp); // Sort the given array sort(a, a + (n - 1)); // Stores the number of times an // edge is part of a path vector< int > ans; // Iterate for all edges and find the // number of nodes on the left and on the right for ( auto it : edges) { // Node 1 int x = it.first; // Node 2 int y = it.second; // If the number of nodes below is less // then the other will be n - dp[node] if (dp[x] < dp[y]) { int fi = dp[x]; int sec = n - dp[x]; ans.push_back(fi * sec); } // Second condition else { int fi = dp[y]; int sec = n - dp[y]; ans.push_back(fi * sec); } } // Sort the number of times // an edges occurs in the path sort(ans.begin(), ans.end()); int res = 0; // Find the summation of all those // paths and return for ( int i = 0; i < n - 1; i++) { res += ans[i] * a[i]; } return res; } // Driver code int main() { int n = 5; vector<pair< int , int > > edges; list< int >* tree = new list< int >[n + 1]; // Add an edge 1-2 in the tree addEdge(edges, tree, 1, 2); // Add an edge 2-3 in the tree addEdge(edges, tree, 1, 3); // Add an edge 3-4 in the tree addEdge(edges, tree, 3, 4); // Add an edge 3-5 in the tree addEdge(edges, tree, 3, 5); // Array which gives the edges weight // to be assigned int a[] = { 6, 3, 1, 9, 3 }; cout << maximizeSum(a, edges, tree, n); } |
Java
import java.util.ArrayList; import java.util.List; class Tree { // Function to add an edge to the tree static void addEdge(List< int []> edges, List<Integer>[] tree, int x, int y) { edges.add( new int [] { x, y }); tree[x].add(y); tree[y].add(x); } // Function to run DFS and calculate the // height of the subtree below it static void dfs(List< int []> edges, List<Integer>[] tree, int node, int parent, int [] dp) { // Initially initialize with 1 dp[node] = 1 ; // Traverse for all nodes connected to node for ( int it : tree[node]) { // If node is not parent // then recall dfs if (it != parent) { dfs(edges, tree, it, node, dp); // Add the size of the // subtree beneath it dp[node] += dp[it]; } } } // Function to assign weights to edges // to maximize the final sum static int maximizeSum( int [] a, List< int []> edges, List<Integer>[] tree, int n) { // Initialize it which stores the // height of the subtree beneath it int [] dp = new int [n + 1 ]; // Call the DFS function to dfs(edges, tree, 1 , 0 , dp); // Sort the given array java.util.Arrays.sort(a, 0 , n - 1 ); // Stores the number of times an // edge is part of a path List<Integer> ans = new ArrayList<Integer>(); // Iterate for all edges and find the // number of nodes on the left and on the right for ( int [] it : edges) { // Node 1 int x = it[ 0 ]; // Node 2 int y = it[ 1 ]; // If the number of nodes below is less // then the other will be n - dp[node] if (dp[x] < dp[y]) { int fi = n - dp[x]; int sec = dp[x]; ans.add(fi * sec); } // Second condition else { int fi = n - dp[y]; int sec = dp[y]; ans.add(fi * sec); } } // Sort the number of times // an edges occurs in the path ans.sort( null ); int res = 0 ; // Find the summation of all those // paths and return for ( int i = 0 ; i < n - 1 ; i++) { res += ans.get(i) * a[i]; } return res; } // Driver code public static void main(String[] args) { int n = 5 ; List< int []> edges = new ArrayList< int []>(); List<Integer>[] tree = new ArrayList[n + 1 ]; for ( int i = 0 ; i < n + 1 ; i++) { tree[i] = new ArrayList<Integer>(); } // Add an edge 1-2 in the tree addEdge(edges, tree, 1 , 2 ); // Add an edge 2-3 in the tree addEdge(edges, tree, 1 , 3 ); // Add an edge 3-4 in the tree addEdge(edges, tree, 3 , 4 ); // Add an edge 3-5 in the tree addEdge(edges, tree, 3 , 5 ); // Array which gives the edges weight // to be assigned int [] a = { 6 , 3 , 1 , 9 , 3 }; System.out.println(maximizeSum(a, edges, tree, n)); } } // This code is contributed by lokeshpotta20. |
Python3
# Python3 program to implement the # above approach edges = [[] for i in range ( 100 )] tree = [[] for i in range ( 100 )] # Function to add an edge to the tree def addEdge(x, y): edges.append([x, y]) tree[x].append(y) tree[y].append(x) # Function to run DFS and calculate the # height of the subtree below it def dfs(node, parent, dp): # Initially initialize with 1 dp[node] = 1 # Traverse for all nodes connected to node for it in tree[node]: # If node is not parent # then recall dfs if (it ! = parent): dfs(it, node, dp) # Add the size of the # subtree beneath it dp[node] + = dp[it] # Function to assign weights to edges # to maximize the final sum def maximizeSum(a, n): # Initialize it which stores the # height of the subtree beneath it dp = [ 0 for i in range (n + 1 )] # Call the DFS function to dfs( 1 , 0 , dp) # Sort the given array a = sorted (a[: - 1 ]) # Stores the number of times an # edge is part of a path ans = [] # Iterate for all edges and find the # number of nodes on the left and on the right for it in edges: if len (it) > 0 : # Node 1 x = it[ 0 ] # Node 2 y = it[ 1 ] # If the number of nodes below is less # then the other will be n - dp[node] if (dp[x] < dp[y]): fi = dp[x] sec = n - dp[x] ans.append(fi * sec) # Second condition else : fi = dp[y] sec = n - dp[y] ans.append(fi * sec) # Sort the number of times # an edges occurs in the path ans = sorted (ans) res = 0 # Find the summation of all those # paths and return for i in range (n - 1 ): res + = ans[i] * a[i] return res # Driver code n = 5 # Add an edge 1-2 in the tree addEdge( 1 , 2 ) # Add an edge 2-3 in the tree addEdge( 1 , 3 ) # Add an edge 3-4 in the tree addEdge( 3 , 4 ) # Add an edge 3-5 in the tree addEdge( 3 , 5 ) # Array which gives the edges weight # to be assigned a = [ 6 , 3 , 1 , 9 , 3 ] print (maximizeSum(a, n)) # This code is contributed by Mohit Kumar |
Javascript
<script> // JavaScript program to implement the // above approach let edges = new Array().fill(0).map(()=> new Array()) let tree = new Array(100).fill(0).map(()=> new Array()) // Function to add an edge to the tree function addEdge(x, y){ edges.push([x, y]) tree[x].push(y) tree[y].push(x) } // Function to run DFS and calculate the // height of the subtree below it function dfs(node, parent, dp){ // Initially initialize with 1 dp[node] = 1 // Traverse for all nodes connected to node for (let it of tree[node]){ // If node is not parent // then recall dfs if (it != parent){ dfs(it, node, dp) // Add the size of the // subtree beneath it dp[node] += dp[it] } } } // Function to assign weights to edges // to maximize the final sum function maximizeSum(a, n){ // Initialize it which stores the // height of the subtree beneath it let dp = new Array(n+1).fill(0) // Call the DFS function to dfs(1, 0, dp) // Sort the given array a = a.slice(0, n-1).sort().concat(a.slice(n-1,)); // Stores the number of times an // edge is part of a path let ans = [] // Iterate for all edges and find the // number of nodes on the left and on the right for (let it of edges){ if (it.length > 0){ // Node 1 let x = it[0] // Node 2 let y = it[1] // If the number of nodes below is less // then the other will be n - dp[node] if (dp[x] < dp[y]){ let fi = dp[x] let sec = n - dp[x] ans.push(fi * sec) } // Second condition else { let fi = dp[y] let sec = n - dp[y] ans.push(fi * sec) } } } // Sort the number of times // an edges occurs in the path ans.sort() let res = 0 // Find the summation of all those // paths and return for (let i=0;i<n-1;i++) res += ans[i] * a[i] return res } // Driver code let n = 5 // Add an edge 1-2 in the tree addEdge(1, 2) // Add an edge 2-3 in the tree addEdge(1, 3) // Add an edge 3-4 in the tree addEdge(3, 4) // Add an edge 3-5 in the tree addEdge(3, 5) // Array which gives the edges weight // to be assigned let a = [6, 3, 1, 9, 3] document.write(maximizeSum(a, n), "</br>" ) // This code is contributed by shinjanpatra </script> |
C#
// C# code for the above approach using System; using System.Collections.Generic; namespace Tree { class Program { // Function to add an edge to the tree static void AddEdge(List<( int , int )> edges, List< int >[] tree, int x, int y) { edges.Add((x, y)); tree[x].Add(y); tree[y].Add(x); } // Function to run DFS and calculate the // height of the subtree below it static void Dfs(List<( int , int )> edges, List< int >[] tree, int node, int parent, int [] dp) { // Initially initialize with 1 dp[node] = 1; // Traverse for all nodes connected to node foreach ( var it in tree[node]) { // If node is not parent // then recall dfs if (it != parent) { Dfs(edges, tree, it, node, dp); // Add the size of the // subtree beneath it dp[node] += dp[it]; } } } // Function to assign weights to edges // to maximize the final sum static int MaximizeSum( int [] a, List<( int , int )> edges, List< int >[] tree, int n) { // Initialize it which stores the // height of the subtree beneath it int [] dp = new int [n + 1]; // Call the DFS function to Dfs(edges, tree, 1, 0, dp); // Sort the given array Array.Sort(a, 0, n - 1); ; // Stores the number of times an // edge is part of a path List< int > ans = new List< int >(); // Iterate for all edges and find the // number of nodes on the left and on the right foreach ( var it in edges) { // Node 1 int x = it.Item1; // Node 2 int y = it.Item2; // If the number of nodes below is less // then the other will be n - dp[node] if (dp[x] < dp[y]) { int fi = n - dp[x]; int sec = dp[x]; ans.Add(fi * sec); } // Second condition else { int fi = n - dp[y]; int sec = dp[y]; ans.Add(fi * sec); } } // Sort the number of times // an edges occurs in the path ans.Sort(); int res = 0; // Find the summation of all those // paths and return for ( int i = 0; i < n - 1; i++) { res += ans[i] * a[i]; } return res; } // Driver code static void Main( string [] args) { int n = 5; List<( int , int )> edges = new List<( int , int )>(); List< int >[] tree = new List< int >[ n + 1 ]; for ( int i = 0; i < n + 1; i++) { tree[i] = new List< int >(); } // Add an edge 1-2 in the tree AddEdge(edges, tree, 1, 2); // Add an edge 2-3 in the tree AddEdge(edges, tree, 1, 3); // Add an edge 3-4 in the tree AddEdge(edges, tree, 3, 4); // Add an edge 3-5 in the tree AddEdge(edges, tree, 3, 5); // Array which gives the edges weight // to be assigned int [] a = { 6, 3, 1, 9, 3 }; Console.WriteLine(MaximizeSum(a, edges, tree, n)); } } } // This code is contributed by Potta Lokesh |
94
Time Complexity: O(V+E) + O(n log n) , for doing the dfs and sorting
Auxiliary Space: O(n), as extra spaces are used
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