Encuentre la distancia entre dos Nodes en el árbol binario dado para consultas Q

Dado un árbol binario que tiene N Nodes y un peso de N-1 aristas. La distancia entre dos Nodes es la suma del peso de los bordes en el camino entre dos Nodes. Cada consulta contiene dos enteros U y V , la tarea es encontrar la distancia entre los Nodes U y V.

Ejemplos: 

Aporte: 
 

Salida: 3 5 12 12 
Explicación: 
Distancia entre Nodes 1 a 3 = peso(1, 3) = 2 
Distancia entre Nodes 2 a 3 = peso(1, 2) + peso(1, 3) = 5 
Distancia entre Nodes 3 a 5 = peso(1, 3) + peso(1, 2) + peso(2, 5) = 12 
Distancia entre Nodes 4 a 5 = peso(4, 2) + peso(2, 5) = 12 
 

Enfoque: La idea es usar LCA en un árbol usando la técnica de elevación binaria .  

• Binary Lifting es un enfoque de programación dinámica en el que precalculamos una array lca[i][j] donde i = [1, n], j = [1, log(n)] y lca[i][j] contiene 2 ^j -ésimo ancestro del Node i. 
  • Para calcular los valores de lca[][], se puede usar la siguiente recursión

• Como calcularemos la array lca[][], también calcularemos la distancia[][] donde la distancia[i][j] contiene la distancia desde el Node i hasta su 2 ^j -ésimo ancestro 
  • Para calcular los valores de dist[][], se puede utilizar la siguiente recursión.

• Después del cálculo previo, encontramos la distancia entre (u, v) como encontramos el antepasado menos común de (u, v).

A continuación se muestra la implementación del enfoque anterior:

// C++ Program to find distance
// between two nodes using LCA
 
#include <bits/stdc++.h>
using namespace std;
 
#define MAX 1000
 
#define log 10 // log2(MAX)
 
// Array to store the level
// of each node
int level[MAX];
 
int lca[MAX][log];
int dist[MAX][log];
 
// Vector to store tree
vector<pair<int, int> > graph[MAX];
 
void addEdge(int u, int v, int cost)
{
    graph[u].push_back({ v, cost });
    graph[v].push_back({ u, cost });
}
 
// Pre-Processing to calculate
// values of lca[][], dist[][]
void dfs(int node, int parent,
         int h, int cost)
{
    // Using recursion formula to
    // calculate the values
    // of lca[][]
    lca[node][0] = parent;
 
    // Storing the level of
    // each node
    level[node] = h;
    if (parent != -1) {
        dist[node][0] = cost;
    }
 
    for (int i = 1; i < log; i++) {
        if (lca[node][i - 1] != -1) {
 
            // Using recursion formula to
            // calculate the values of
            // lca[][] and dist[][]
            lca[node][i]
                = lca[lca[node]
                         [i - 1]]
                     [i - 1];
 
            dist[node][i]
                = dist[node][i - 1]
                  + dist[lca[node][i - 1]]
                        [i - 1];
        }
    }
 
    for (auto i : graph[node]) {
        if (i.first == parent)
            continue;
        dfs(i.first, node,
h + 1, i.second);
    }
}
 
// Function to find the distance
// between given nodes u and v
void findDistance(int u, int v)
{
 
    int ans = 0;
 
    // The node which is present
    // farthest from the root node
    // is taken as v. If u is
    // farther from root node
    // then swap the two
    if (level[u] > level[v])
        swap(u, v);
 
    // Finding the ancestor of v
    // which is at same level as u
    for (int i = log - 1; i >= 0; i--) {
 
        if (lca[v][i] != -1
            && level[lca[v][i]]
                   >= level[u]) {
 
            // Adding distance of node
            // v till its 2^i-th ancestor
            ans += dist[v][i];
            v = lca[v][i];
        }
    }
 
    // If u is the ancestor of v
    // then u is the LCA of u and v
    if (v == u) {
 
        cout << ans << endl;
    }
 
    else {
 
        // Finding the node closest to the
        // root which is not the common
        // ancestor of u and v i.e. a node
        // x such that x is not the common
        // ancestor of u and v but lca[x][0] is
        for (int i = log - 1; i >= 0; i--) {
 
            if (lca[v][i] != lca[u][i]) {
 
                // Adding the distance
                // of v and u to
                // its 2^i-th ancestor
                ans += dist[u][i] + dist[v][i];
 
                v = lca[v][i];
                u = lca[u][i];
            }
        }
 
        // Adding the distance of u and v
        // to its first ancestor
        ans += dist[u][0] + dist[v][0];
 
        cout << ans << endl;
    }
}
 
// Driver Code
int main()
{
 
    // Number of nodes
    int n = 5;
 
    // Add edges with their cost
    addEdge(1, 2, 2);
    addEdge(1, 3, 3);
    addEdge(2, 4, 5);
    addEdge(2, 5, 7);
 
    // Initialising lca and dist values
    // with -1 and 0 respectively
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j < log; j++) {
            lca[i][j] = -1;
            dist[i][j] = 0;
        }
    }
 
    // Perform DFS
    dfs(1, -1, 0, 0);
 
    // Query 1: {1, 3}
    findDistance(1, 3);
 
    // Query 2: {2, 3}
    findDistance(2, 3);
 
    // Query 3: {3, 5}
    findDistance(3, 5);
 
    return 0;
}

// Java program to find distance
// between two nodes using LCA
import java.io.*;
import java.util.*;
 
class GFG{
 
static final int MAX = 1000;
// log2(MAX)
static final int log = 10;
 
// Array to store the level
// of each node
static int[] level = new int[MAX];
 
static int[][] lca = new int[MAX][log];
static int[][] dist = new int[MAX][log];
 
// Vector to store tree
@SuppressWarnings("unchecked")
static List<List<int[]> > graph = new ArrayList();
 
static void addEdge(int u, int v, int cost)
{
    graph.get(u).add(new int[]{ v, cost });
    graph.get(v).add(new int[]{ u, cost });
}
 
// Pre-Processing to calculate
// values of lca[][], dist[][]
static void dfs(int node, int parent,
                int h, int cost)
{
     
    // Using recursion formula to
    // calculate the values
    // of lca[][]
    lca[node][0] = parent;
 
    // Storing the level of
    // each node
    level[node] = h;
     
    if (parent != -1)
    {
        dist[node][0] = cost;
    }
 
    for(int i = 1; i < log; i++)
    {
        if (lca[node][i - 1] != -1)
        {
             
            // Using recursion formula to
            // calculate the values of
            // lca[][] and dist[][]
            lca[node][i] = lca[lca[node][i - 1]][i - 1];
 
            dist[node][i] = dist[node][i - 1] +
                            dist[lca[node][i - 1]][i - 1];
        }
    }
 
    for(int[] i : graph.get(node))
    {
        if (i[0] == parent)
            continue;
             
        dfs(i[0], node, h + 1, i[1]);
    }
}
 
// Function to find the distance
// between given nodes u and v
static void findDistance(int u, int v)
{
    int ans = 0;
 
    // The node which is present
    // farthest from the root node
    // is taken as v. If u is
    // farther from root node
    // then swap the two
    if (level[u] > level[v])
    {
        int temp = u;
        u = v;
        v = temp;
    }
 
    // Finding the ancestor of v
    // which is at same level as u
    for(int i = log - 1; i >= 0; i--)
    {
        if (lca[v][i] != -1 &&
      level[lca[v][i]] >= level[u])
        {
 
            // Adding distance of node
            // v till its 2^i-th ancestor
            ans += dist[v][i];
            v = lca[v][i];
        }
    }
 
    // If u is the ancestor of v
    // then u is the LCA of u and v
    if (v == u)
    {
        System.out.println(ans);
    }
 
    else
    {
         
        // Finding the node closest to the
        // root which is not the common
        // ancestor of u and v i.e. a node
        // x such that x is not the common
        // ancestor of u and v but lca[x][0] is
        for(int i = log - 1; i >= 0; i--)
        {
            if (lca[v][i] != lca[u][i])
            {
                 
                // Adding the distance
                // of v and u to
                // its 2^i-th ancestor
                ans += dist[u][i] + dist[v][i];
 
                v = lca[v][i];
                u = lca[u][i];
            }
        }
 
        // Adding the distance of u and v
        // to its first ancestor
        ans += dist[u][0] + dist[v][0];
 
        System.out.println(ans);
    }
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Number of nodes
    int n = 5;
 
    for(int i = 0; i < MAX; i++)
    {
        graph.add(new ArrayList<int[]>());
    }
 
    // Add edges with their cost
    addEdge(1, 2, 2);
    addEdge(1, 3, 3);
    addEdge(2, 4, 5);
    addEdge(2, 5, 7);
 
    // Initialising lca and dist values
    // with -1 and 0 respectively
    for(int i = 1; i <= n; i++)
    {
        for(int j = 0; j < log; j++)
        {
            lca[i][j] = -1;
            dist[i][j] = 0;
        }
    }
 
    // Perform DFS
    dfs(1, -1, 0, 0);
 
    // Query 1: {1, 3}
    findDistance(1, 3);
 
    // Query 2: {2, 3}
    findDistance(2, 3);
 
    // Query 3: {3, 5}
    findDistance(3, 5);
}
}
 
// This code is contributed by jithin

# Python3 Program to find
# distance between two nodes
# using LCA
MAX = 1000
 
# lg2(MAX)
lg = 10
 
# Array to store the level
# of each node
level = [0 for i in range(MAX)]
 
lca = [[0 for i in range(lg)]
          for j in range(MAX)]
dist = [[0 for i in range(lg)]
           for j in range(MAX)]
 
# Vector to store tree
graph = [[] for i in range(MAX)]
 
def addEdge(u, v, cost):
   
    global graph
     
    graph[u].append([v, cost])
    graph[v].append([u, cost])
 
# Pre-Processing to calculate
# values of lca[][], dist[][]
def dfs(node, parent, h, cost):
   
    # Using recursion formula to
    # calculate the values
    # of lca[][]
    lca[node][0] = parent
 
    # Storing the level of
    # each node
    level[node] = h
     
    if (parent != -1):
        dist[node][0] = cost
 
    for i in range(1, lg):
        if (lca[node][i - 1] != -1):
           
            # Using recursion formula to
            # calculate the values of
            # lca[][] and dist[][]
            lca[node][i] = lca[lca[node][i - 1]][i - 1]
 
            dist[node][i] = (dist[node][i - 1] +
                             dist[lca[node][i - 1]][i - 1])
 
    for i in graph[node]:
        if (i[0] == parent):
            continue
        dfs(i[0], node, h + 1, i[1])
 
# Function to find the distance
# between given nodes u and v
def findDistance(u, v):
   
    ans = 0
 
    # The node which is present
    # farthest from the root node
    # is taken as v. If u is
    # farther from root node
    # then swap the two
    if (level[u] > level[v]):
        temp = u
        u = v
        v = temp
 
    # Finding the ancestor of v
    # which is at same level as u
    i = lg - 1
     
    while(i >= 0):
        if (lca[v][i] != -1 and
            level[lca[v][i]] >= level[u]):
           
            # Adding distance of node
            # v till its 2^i-th ancestor
            ans += dist[v][i]
            v = lca[v][i]
             
        i -= 1
 
    # If u is the ancestor of v
    # then u is the LCA of u and v
    if (v == u):
        print(ans)
 
    else:
        # Finding the node closest to the
        # root which is not the common
        # ancestor of u and v i.e. a node
        # x such that x is not the common
        # ancestor of u and v but lca[x][0] is
        i = lg - 1
         
        while(i >= 0):
            if (lca[v][i] != lca[u][i]):
                # Adding the distance
                # of v and u to
                # its 2^i-th ancestor
                ans += dist[u][i] + dist[v][i]
 
                v = lca[v][i]
                u = lca[u][i]
            i -= 1
 
        # Adding the distance of u and v
        # to its first ancestor
        ans += (dist[u][0] +
                dist[v][0])
 
        print(ans)
 
# Driver Code
if __name__ == '__main__':
   
    # Number of nodes
    n = 5
 
    # Add edges with their cost
    addEdge(1, 2, 2)
    addEdge(1, 3, 3)
    addEdge(2, 4, 5)
    addEdge(2, 5, 7)
 
    # Initialising lca and dist values
    # with -1 and 0 respectively
    for i in range(1, n + 1):
        for j in range(lg):
            lca[i][j] = -1
            dist[i][j] = 0
             
    # Perform DFS
    dfs(1, -1, 0, 0)
    # Query 1: {1, 3}
    findDistance(1, 3)
    # Query 2: {2, 3}
    findDistance(2, 3)
    # Query 3: {3, 5}
    findDistance(3, 5)
     
# This code is contributed by SURENDRA_GANGWAR

// C# program to find distance
// between two nodes using LCA
using System;
using System.Collections.Generic;
class GFG
{
 
  static readonly int MAX = 1000;
 
  // log2(MAX)
  static readonly int log = 10;
 
  // Array to store the level
  // of each node
  static int[] level = new int[MAX];
  static int[,] lca = new int[MAX,log];
  static int[,] dist = new int[MAX,log];
 
  // List to store tree
  static List<List<int[]> > graph = new List<List<int[]>>();
 
  static void addEdge(int u, int v, int cost)
  {
    graph[u].Add(new int[]{ v, cost });
    graph[v].Add(new int[]{ u, cost });
  }
 
  // Pre-Processing to calculate
  // values of lca[,], dist[,]
  static void dfs(int node, int parent,
                  int h, int cost)
  {
 
    // Using recursion formula to
    // calculate the values
    // of lca[,]
    lca[node, 0] = parent;
 
    // Storing the level of
    // each node
    level[node] = h;
 
    if (parent != -1)
    {
      dist[node, 0] = cost;
    }
 
    for(int i = 1; i < log; i++)
    {
      if (lca[node, i - 1] != -1)
      {
 
        // Using recursion formula to
        // calculate the values of
        // lca[,] and dist[,]
        lca[node, i] = lca[lca[node, i - 1], i - 1];
 
        dist[node, i] = dist[node, i - 1] +
          dist[lca[node, i - 1], i - 1];
      }
    }
 
    foreach(int[] i in graph[node])
    {
      if (i[0] == parent)
        continue;           
      dfs(i[0], node, h + 1, i[1]);
    }
  }
 
  // Function to find the distance
  // between given nodes u and v
  static void findDistance(int u, int v)
  {
    int ans = 0;
 
    // The node which is present
    // farthest from the root node
    // is taken as v. If u is
    // farther from root node
    // then swap the two
    if (level[u] > level[v])
    {
      int temp = u;
      u = v;
      v = temp;
    }
 
    // Finding the ancestor of v
    // which is at same level as u
    for(int i = log - 1; i >= 0; i--)
    {
      if (lca[v, i] != -1 &&
          level[lca[v, i]] >= level[u])
      {
 
        // Adding distance of node
        // v till its 2^i-th ancestor
        ans += dist[v, i];
        v = lca[v, i];
      }
    }
 
    // If u is the ancestor of v
    // then u is the LCA of u and v
    if (v == u)
    {
      Console.WriteLine(ans);
    }
 
    else
    {
 
      // Finding the node closest to the
      // root which is not the common
      // ancestor of u and v i.e. a node
      // x such that x is not the common
      // ancestor of u and v but lca[x,0] is
      for(int i = log - 1; i >= 0; i--)
      {
        if (lca[v, i] != lca[u, i])
        {
 
          // Adding the distance
          // of v and u to
          // its 2^i-th ancestor
          ans += dist[u, i] + dist[v, i];
 
          v = lca[v, i];
          u = lca[u, i];
        }
      }
 
      // Adding the distance of u and v
      // to its first ancestor
      ans += dist[u, 0] + dist[v, 0];
      Console.WriteLine(ans);
    }
  }
 
  // Driver Code
  public static void Main(String[] args)
  {
 
    // Number of nodes
    int n = 5;
    for(int i = 0; i < MAX; i++)
    {
      graph.Add(new List<int[]>());
    }
 
    // Add edges with their cost
    addEdge(1, 2, 2);
    addEdge(1, 3, 3);
    addEdge(2, 4, 5);
    addEdge(2, 5, 7);
 
    // Initialising lca and dist values
    // with -1 and 0 respectively
    for(int i = 1; i <= n; i++)
    {
      for(int j = 0; j < log; j++)
      {
        lca[i, j] = -1;
        dist[i, j] = 0;
      }
    }
 
    // Perform DFS
    dfs(1, -1, 0, 0);
 
    // Query 1: {1, 3}
    findDistance(1, 3);
 
    // Query 2: {2, 3}
    findDistance(2, 3);
 
    // Query 3: {3, 5}
    findDistance(3, 5);
  }
}
 
// This code is contributed by aashish1995

<script>
    // Javascript program to find distance
    // between two nodes using LCA
     
    let MAX = 1000;
    // log2(MAX)
    let log = 10;
 
    // Array to store the level
    // of each node
    let level = new Array(MAX);
 
    let lca = new Array(MAX);
    let dist = new Array(MAX);
 
    // Vector to store tree
    let graph = [];
 
    function addEdge(u, v, cost)
    {
        graph[u].push([ v, cost ]);
        graph[v].push([ u, cost ]);
    }
 
    // Pre-Processing to calculate
    // values of lca[][], dist[][]
    function dfs(node, parent, h, cost)
    {
 
        // Using recursion formula to
        // calculate the values
        // of lca[][]
        lca[node][0] = parent;
 
        // Storing the level of
        // each node
        level[node] = h;
 
        if (parent != -1)
        {
            dist[node][0] = cost;
        }
 
        for(let i = 1; i < log; i++)
        {
            if (lca[node][i - 1] != -1)
            {
 
                // Using recursion formula to
                // calculate the values of
                // lca[][] and dist[][]
                lca[node][i] = lca[lca[node][i - 1]][i - 1];
 
                dist[node][i] = dist[node][i - 1] +
                                dist[lca[node][i - 1]][i - 1];
            }
        }
 
        for(let i = 0; i < graph[node].length; i++)
        {
            if (graph[node][i][0] == parent)
                continue;
 
            dfs(graph[node][i][0], node, h + 1, graph[node][i][1]);
        }
    }
 
    // Function to find the distance
    // between given nodes u and v
    function findDistance(u, v)
    {
        let ans = 0;
 
        // The node which is present
        // farthest from the root node
        // is taken as v. If u is
        // farther from root node
        // then swap the two
        if (level[u] > level[v])
        {
            let temp = u;
            u = v;
            v = temp;
        }
 
        // Finding the ancestor of v
        // which is at same level as u
        for(let i = log - 1; i >= 0; i--)
        {
            if (lca[v][i] != -1 && level[lca[v][i]] >= level[u])
            {
 
                // Adding distance of node
                // v till its 2^i-th ancestor
                ans += dist[v][i];
                v = lca[v][i];
            }
        }
 
        // If u is the ancestor of v
        // then u is the LCA of u and v
        if (v == u)
        {
            document.write(ans + "</br>");
        }
 
        else
        {
 
            // Finding the node closest to the
            // root which is not the common
            // ancestor of u and v i.e. a node
            // x such that x is not the common
            // ancestor of u and v but lca[x][0] is
            for(let i = log - 1; i >= 0; i--)
            {
                if (lca[v][i] != lca[u][i])
                {
 
                    // Adding the distance
                    // of v and u to
                    // its 2^i-th ancestor
                    ans += dist[u][i] + dist[v][i];
 
                    v = lca[v][i];
                    u = lca[u][i];
                }
            }
 
            // Adding the distance of u and v
            // to its first ancestor
            ans += dist[u][0] + dist[v][0];
 
            document.write(ans + "</br>");
        }
    }
     
    // Number of nodes
    let n = 5;
  
    for(let i = 0; i < MAX; i++)
    {
        graph.push([]);
    }
  
    // Add edges with their cost
    addEdge(1, 2, 2);
    addEdge(1, 3, 3);
    addEdge(2, 4, 5);
    addEdge(2, 5, 7);
  
    // Initialising lca and dist values
    // with -1 and 0 respectively
    for(let i = 1; i <= n; i++)
    {
        lca[i] = new Array(log);
        dist[i] = new Array(log);
        for(let j = 0; j < log; j++)
        {
            lca[i][j] = -1;
            dist[i][j] = 0;
        }
    }
  
    // Perform DFS
    dfs(1, -1, 0, 0);
  
    // Query 1: {1, 3}
    findDistance(1, 3);
  
    // Query 2: {2, 3}
    findDistance(2, 3);
  
    // Query 3: {3, 5}
    findDistance(3, 5);
 
// This code is contributed by decode2207.
</script>

3
5
12

Complejidad del tiempo: el tiempo necesario para el preprocesamiento es O (N logN) y cada consulta lleva un tiempo O (logN) . Por lo tanto, la complejidad temporal total de la solución es O(N logN) .