Hem vist diversos mètodes amb diferents complexitats temporals per calcular l'ACV a l'arbre n-ari: -
Mètode 1: Mètode ingenu (calculant el camí de l'arrel al node) | O(n) per consulta
Mètode 2: Utilitzant la descomposició Sqrt | O (h quadrat)
Mètode 3: Utilitzant l'enfocament Sparse Matrix DP | O (inici de sessió)
Estudiem un altre mètode que té un temps de consulta més ràpid que tots els mètodes anteriors. Per tant, el nostre objectiu serà calcular l'LCA temps constant ~ O(1) . A veure com ho podem aconseguir.
Mètode 4: ús de la consulta mínima d'interval
Hem discutit LCA i RMQ per a arbre binari . Aquí discutim la conversió del problema LCA a problema RMQ per a l'arbre n-ari.
Pre-requisites:- LCA in Binary Tree using RMQ RMQ using sparse table
Concepte clau: En aquest mètode reduirem el nostre problema LCA a un problema RMQ (Consulta mínima de rang) sobre una matriu estàtica. Un cop ho fem, relacionarem les consultes mínimes d'interval amb les consultes LCA necessàries.
El primer pas serà descompondre l'arbre en una matriu lineal plana. Per fer-ho podem aplicar la caminada d'Euler. El passeig d'Euler donarà el recorregut de pre-ordre del gràfic. Per tant, realitzarem un Euler Walk a l'arbre i emmagatzemarem els nodes en una matriu mentre els visitem. Aquest procés redueix l'arbre > 
Ara pensem en termes generals: considerem dos nodes qualsevol de l'arbre. Hi haurà exactament un camí que connecti els dos nodes i el node que tingui el valor de profunditat més petit del camí serà l'LCA dels dos nodes donats.
Ara prengueu dos nodes diferents, per exemple en i v a la matriu de passeig d'Euler. Ara tots els elements del camí d'u a v estaran entre l'índex dels nodes u i v de la matriu de caminada d'Euler. Per tant, només hem de calcular el node amb la profunditat mínima entre l'índex del node u i el node v a la matriu d'Euler.
Per a això mantindrem una altra matriu que contindrà la profunditat de tots els nodes corresponents a la seva posició en la matriu de caminada d'Euler perquè puguem aplicar-hi el nostre algorisme RMQ.
A continuació es mostra la matriu Euler Walk paral·lela a la seva matriu de pistes de profunditat.
per cada java
Exemple: - Considereu dos nodes node 6 i node 7 a la matriu euler. Per calcular l'LCA del node 6 i del node 7 mirem el valor de profunditat més petit per a tots els nodes entre el node 6 i el node 7.
Per tant node 1 té el més petit valor de profunditat = 0 i per tant és l'LCA per al node 6 i el node 7.
Implementació: -
We will be maintaining three arrays 1) Euler Path 2) Depth array 3) First Appearance Index
Euler Path i Depth matriu són els mateixos que s'han descrit anteriorment
Índex de primera aparició FAI[] : La matriu d'índex de First Appearance emmagatzemarà l'índex de la primera posició de cada node de la matriu d'Euler Path. FAI[i] = Primera aparició del node iè a la matriu Euler Walk.
A continuació es detalla la implementació del mètode anterior:
Implementació:
corda al cC++
// C++ program to demonstrate LCA of n-ary tree // in constant time. #include 'bits/stdc++.h' using namespace std; #define sz 101 vector < int > adj[sz]; // stores the tree vector < int > euler; // tracks the eulerwalk vector < int > depthArr; // depth for each node corresponding // to eulerwalk int FAI[sz]; // stores first appearance index of every node int level[sz]; // stores depth for all nodes in the tree int ptr; // pointer to euler walk int dp[sz][18]; // sparse table int logn[sz]; // stores log values int p2[20]; // stores power of 2 void buildSparseTable(int n) { // initializing sparse table memset(dp-1sizeof(dp)); // filling base case values for (int i=1; i<n; i++) dp[i-1][0] = (depthArr[i]>depthArr[i-1])?i-1:i; // dp to fill sparse table for (int l=1; l<15; l++) for (int i=0; i<n; i++) if (dp[i][l-1]!=-1 and dp[i+p2[l-1]][l-1]!=-1) dp[i][l] = (depthArr[dp[i][l-1]]>depthArr[dp[i+p2[l-1]][l-1]])? dp[i+p2[l-1]][l-1] : dp[i][l-1]; else break; } int query(int lint r) { int d = r-l; int dx = logn[d]; if (l==r) return l; if (depthArr[dp[l][dx]] > depthArr[dp[r-p2[dx]][dx]]) return dp[r-p2[dx]][dx]; else return dp[l][dx]; } void preprocess() { // memorizing powers of 2 p2[0] = 1; for (int i=1; i<18; i++) p2[i] = p2[i-1]*2; // memorizing all log(n) values int val = 1ptr=0; for (int i=1; i<sz; i++) { logn[i] = ptr-1; if (val==i) { val*=2; logn[i] = ptr; ptr++; } } } /** * Euler Walk ( preorder traversal) * converting tree to linear depthArray * Time Complexity : O(n) * */ void dfs(int curint prevint dep) { // marking FAI for cur node if (FAI[cur]==-1) FAI[cur] = ptr; level[cur] = dep; // pushing root to euler walk euler.push_back(cur); // incrementing euler walk pointer ptr++; for (auto x:adj[cur]) { if (x != prev) { dfs(xcurdep+1); // pushing cur again in backtrack // of euler walk euler.push_back(cur); // increment euler walk pointer ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array void makeArr() { for (auto x : euler) depthArr.push_back(level[x]); } int LCA(int uint v) { // trivial case if (u==v) return u; if (FAI[u] > FAI[v]) swap(uv); // doing RMQ in the required range return euler[query(FAI[u] FAI[v])]; } void addEdge(int uint v) { adj[u].push_back(v); adj[v].push_back(u); } int main(int argc char const *argv[]) { // constructing the described tree int numberOfNodes = 8; addEdge(12); addEdge(13); addEdge(24); addEdge(25); addEdge(26); addEdge(37); addEdge(38); // performing required precalculations preprocess(); // doing the Euler walk ptr = 0; memset(FAI-1sizeof(FAI)); dfs(100); // creating depthArray corresponding to euler[] makeArr(); // building sparse table buildSparseTable(depthArr.size()); cout << 'LCA(67) : ' << LCA(67) << 'n'; cout << 'LCA(64) : ' << LCA(64) << 'n'; return 0; }
// Java program to demonstrate LCA of n-ary // tree in constant time. import java.util.ArrayList; import java.util.Arrays; class GFG{ static int sz = 101; @SuppressWarnings('unchecked') // Stores the tree static ArrayList<Integer>[] adj = new ArrayList[sz]; // Tracks the eulerwalk static ArrayList<Integer> euler = new ArrayList<>(); // Depth for each node corresponding static ArrayList<Integer> depthArr = new ArrayList<>(); // to eulerwalk // Stores first appearance index of every node static int[] FAI = new int[sz]; // Stores depth for all nodes in the tree static int[] level = new int[sz]; // Pointer to euler walk static int ptr; // Sparse table static int[][] dp = new int[sz][18]; // Stores log values static int[] logn = new int[sz]; // Stores power of 2 static int[] p2 = new int[20]; static void buildSparseTable(int n) { // Initializing sparse table for(int i = 0; i < sz; i++) { for(int j = 0; j < 18; j++) { dp[i][j] = -1; } } // Filling base case values for(int i = 1; i < n; i++) dp[i - 1][0] = (depthArr.get(i) > depthArr.get(i - 1)) ? i - 1 : i; // dp to fill sparse table for(int l = 1; l < 15; l++) for(int i = 0; i < n; i++) if (dp[i][l - 1] != -1 && dp[i + p2[l - 1]][l - 1] != -1) dp[i][l] = (depthArr.get(dp[i][l - 1]) > depthArr.get( dp[i + p2[l - 1]][l - 1])) ? dp[i + p2[l - 1]][l - 1] : dp[i][l - 1]; else break; } static int query(int l int r) { int d = r - l; int dx = logn[d]; if (l == r) return l; if (depthArr.get(dp[l][dx]) > depthArr.get(dp[r - p2[dx]][dx])) return dp[r - p2[dx]][dx]; else return dp[l][dx]; } static void preprocess() { // Memorizing powers of 2 p2[0] = 1; for(int i = 1; i < 18; i++) p2[i] = p2[i - 1] * 2; // Memorizing all log(n) values int val = 1 ptr = 0; for(int i = 1; i < sz; i++) { logn[i] = ptr - 1; if (val == i) { val *= 2; logn[i] = ptr; ptr++; } } } // Euler Walk ( preorder traversal) converting // tree to linear depthArray // Time Complexity : O(n) static void dfs(int cur int prev int dep) { // Marking FAI for cur node if (FAI[cur] == -1) FAI[cur] = ptr; level[cur] = dep; // Pushing root to euler walk euler.add(cur); // Incrementing euler walk pointer ptr++; for(Integer x : adj[cur]) { if (x != prev) { dfs(x cur dep + 1); // Pushing cur again in backtrack // of euler walk euler.add(cur); // Increment euler walk pointer ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array static void makeArr() { for(Integer x : euler) depthArr.add(level[x]); } static int LCA(int u int v) { // Trivial case if (u == v) return u; if (FAI[u] > FAI[v]) { int temp = u; u = v; v = temp; } // Doing RMQ in the required range return euler.get(query(FAI[u] FAI[v])); } static void addEdge(int u int v) { adj[u].add(v); adj[v].add(u); } // Driver code public static void main(String[] args) { for(int i = 0; i < sz; i++) { adj[i] = new ArrayList<>(); } // Constructing the described tree int numberOfNodes = 8; addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); // Performing required precalculations preprocess(); // Doing the Euler walk ptr = 0; Arrays.fill(FAI -1); dfs(1 0 0); // Creating depthArray corresponding to euler[] makeArr(); // Building sparse table buildSparseTable(depthArr.size()); System.out.println('LCA(67) : ' + LCA(6 7)); System.out.println('LCA(64) : ' + LCA(6 4)); } } // This code is contributed by sanjeev2552
# Python program to demonstrate LCA of n-ary tree # in constant time. from typing import List # stores the tree adj = [[] for _ in range(101)] # tracks the eulerwalk euler = [] # depth for each node corresponding to eulerwalk depthArr = [] # stores first appearance index of every node FAI = [-1] * 101 # stores depth for all nodes in the tree level = [0] * 101 # pointer to euler walk ptr = 0 # sparse table dp = [[-1] * 18 for _ in range(101)] # stores log values logn = [0] * 101 # stores power of 2 p2 = [0] * 20 def buildSparseTable(n: int): # initializing sparse table for i in range(n): dp[i][0] = i-1 if depthArr[i] > depthArr[i-1] else i # dp to fill sparse table for l in range(1 15): for i in range(n): if dp[i][l-1] != -1 and dp[i+p2[l-1]][l-1] != -1: dp[i][l] = dp[i+p2[l-1]][l-1] if depthArr[dp[i][l-1] ] > depthArr[dp[i+p2[l-1]][l-1]] else dp[i][l-1] else: break def query(l: int r: int) -> int: d = r-l dx = logn[d] if l == r: return l if depthArr[dp[l][dx]] > depthArr[dp[r-p2[dx]][dx]]: return dp[r-p2[dx]][dx] else: return dp[l][dx] def preprocess(): global ptr # memorizing powers of 2 p2[0] = 1 for i in range(1 18): p2[i] = p2[i-1]*2 # memorizing all log(n) values val = 1 ptr = 0 for i in range(1 101): logn[i] = ptr-1 if val == i: val *= 2 logn[i] = ptr ptr += 1 def dfs(cur: int prev: int dep: int): global ptr # marking FAI for cur node if FAI[cur] == -1: FAI[cur] = ptr level[cur] = dep # pushing root to euler walk euler.append(cur) # incrementing euler walk pointer ptr += 1 for x in adj[cur]: if x != prev: dfs(x cur dep+1) # pushing cur again in backtrack # of euler walk euler.append(cur) # increment euler walk pointer ptr += 1 # Create Level depthArray corresponding # to the Euler walk Array def makeArr(): global depthArr for x in euler: depthArr.append(level[x]) def LCA(u: int v: int) -> int: # trivial case if u == v: return u if FAI[u] > FAI[v]: u v = v u # doing RMQ in the required range return euler[query(FAI[u] FAI[v])] def addEdge(u v): adj[u].append(v) adj[v].append(u) # constructing the described tree numberOfNodes = 8 addEdge(1 2) addEdge(1 3) addEdge(2 4) addEdge(2 5) addEdge(2 6) addEdge(3 7) addEdge(3 8) # performing required precalculations preprocess() # doing the Euler walk ptr = 0 FAI = [-1] * (numberOfNodes + 1) dfs(1 0 0) # creating depthArray corresponding to euler[] makeArr() # building sparse table buildSparseTable(len(depthArr)) print('LCA(67) : ' LCA(6 7)) print('LCA(64) : ' LCA(6 4))
// C# program to demonstrate LCA of n-ary // tree in constant time. using System; using System.Collections.Generic; public class GFG { static int sz = 101; // Stores the tree static List<int>[] adj = new List<int>[sz]; // Tracks the eulerwalk static List<int> euler = new List<int>(); // Depth for each node corresponding static List<int> depthArr = new List<int>(); // to eulerwalk // Stores first appearance index of every node static int[] FAI = new int[sz]; // Stores depth for all nodes in the tree static int[] level = new int[sz]; // Pointer to euler walk static int ptr; // Sparse table static int[] dp = new int[sz 18]; // Stores log values static int[] logn = new int[sz]; // Stores power of 2 static int[] p2 = new int[20]; static void buildSparseTable(int n) { // Initializing sparse table for(int i = 0; i < sz; i++) { for(int j = 0; j < 18; j++) { dp[ij] = -1; } } // Filling base case values for(int i = 1; i < n; i++) dp[i - 10] = (depthArr[i] > depthArr[i - 1]) ? i - 1 : i; // dp to fill sparse table for(int l = 1; l < 15; l++) for(int i = 0; i < n; i++) if (dp[il - 1] != -1 && dp[i + p2[l - 1]l - 1] != -1) dp[il] = (depthArr[dp[il - 1]] > depthArr[dp[i + p2[l - 1]l - 1]]) ? dp[i + p2[l - 1]l - 1] : dp[il - 1]; else break; } static int query(int l int r) { int d = r - l; int dx = logn[d]; if (l == r) return l; if (depthArr[dp[ldx]] > depthArr[dp[r - p2[dx]dx]]) return dp[r - p2[dx]dx]; else return dp[ldx]; } static void preprocess() { // Memorizing powers of 2 p2[0] = 1; for(int i = 1; i < 18; i++) p2[i] = p2[i - 1] * 2; // Memorizing all log(n) values int val = 1 ptr = 0; for(int i = 1; i < sz; i++) { logn[i] = ptr - 1; if (val == i) { val *= 2; logn[i] = ptr; ptr++; } } } // Euler Walk ( preorder traversal) converting // tree to linear depthArray // Time Complexity : O(n) static void dfs(int cur int prev int dep) { // Marking FAI for cur node if (FAI[cur] == -1) FAI[cur] = ptr; level[cur] = dep; // Pushing root to euler walk euler.Add(cur); // Incrementing euler walk pointer ptr++; foreach (int x in adj[cur]) { if (x != prev) { dfs(x cur dep + 1); euler.Add(cur); ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array static void makeArr() { foreach (int x in euler) depthArr.Add(level[x]); } static int LCA(int u int v) { // Trivial case if (u == v) return u; if (FAI[u] > FAI[v]) { int temp = u; u = v; v = temp; } // Doing RMQ in the required range return euler[query(FAI[u] FAI[v])]; } static void addEdge(int u int v) { adj[u].Add(v); adj[v].Add(u); } // Driver Code static void Main(string[] args) { int sz = 9; adj = new List<int>[sz]; for (int i = 0; i < sz; i++) { adj[i] = new List<int>(); } // Constructing the described tree int numberOfNodes = 8; addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); // Performing required precalculations preprocess(); // Doing the Euler walk ptr = 0; Array.Fill(FAI -1); dfs(1 0 0); // Creating depthArray corresponding to euler[] makeArr(); // Building sparse table buildSparseTable(depthArr.Count); Console.WriteLine('LCA(67) : ' + LCA(6 7)); Console.WriteLine('LCA(64) : ' + LCA(6 4)); } } // This code is contributed by Prince Kumar
let adj = []; for (let _ = 0; _ < 101; _++) { adj.push([]); } // tracks the eulerwalk let euler = []; // depth for each node corresponding to eulerwalk let depthArr = []; // stores first appearance index of every node let FAI = new Array(101).fill(-1); // stores depth for all nodes in the tree let level = new Array(101).fill(0); // pointer to euler walk let ptr = 0; // sparse table let dp = []; for (let _ = 0; _ < 101; _++) { dp.push(new Array(18).fill(-1)); } // stores log values let logn = new Array(101).fill(0); // stores power of 2 let p2 = new Array(20).fill(0); function buildSparseTable(n) { // initializing sparse table for (let i = 0; i < n; i++) { dp[i][0] = i - 1 >= 0 && depthArr[i] > depthArr[i - 1] ? i - 1 : i; } // dp to fill sparse table for (let l = 1; l < 15; l++) { for (let i = 0; i < n; i++) { if ( dp[i][l - 1] !== -1 && dp[i + p2[l - 1]][l - 1] !== -1 ) { dp[i][l] = depthArr[dp[i][l - 1]] > depthArr[dp[i + p2[l - 1]][l - 1]] ? dp[i + p2[l - 1]][l - 1] : dp[i][l - 1]; } else { break; } } } } function query(l r) { let d = r - l; let dx = logn[d]; if (l === r) { return l; } if (depthArr[dp[l][dx]] > depthArr[dp[r - p2[dx]][dx]]) { return dp[r - p2[dx]][dx]; } else { return dp[l][dx]; } } function preprocess() { // memorizing powers of 2 p2[0] = 1; for (let i = 1; i < 18; i++) { p2[i] = p2[i - 1] * 2; } // memorizing all log(n) values let val = 1; ptr = 0; for (let i = 1; i < 101; i++) { logn[i] = ptr - 1; if (val === i) { val *= 2; logn[i] = ptr; ptr += 1; } } } function dfs(cur prev dep) { // marking FAI for cur node if (FAI[cur] === -1) { FAI[cur] = ptr; } level[cur] = dep; // pushing root to euler walk euler.push(cur); // incrementing euler walk pointer ptr += 1; for (let x of adj[cur]) { if (x !== prev) { dfs(x cur dep + 1); // pushing cur again in backtrack // of euler walk euler.push(cur); // increment euler walk pointer ptr += 1; } } } // Create Level depthArray corresponding // to the Euler walk Array function makeArr() { for (let x of euler) { depthArr.push(level[x]); } } function LCA(u v) { // trivial case if (u === v) { return u; } if (FAI[u] > FAI[v]) { [u v] = [v u]; } // doing RMQ in the required range return euler[query(FAI[u] FAI[v])]; } function addEdge(u v) { adj[u].push(v); adj[v].push(u); } // constructing the described tree let numberOfNodes = 8; addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); // performing required precalculations preprocess(); // doing the Euler walk ptr = 0; FAI = new Array(numberOfNodes + 1).fill(-1); dfs(1 0 0); // creating depthArray corresponding to euler[] makeArr(); // building sparse table buildSparseTable(depthArr.length); console.log('LCA(67) : ' LCA(6 7)); console.log('LCA(64) : ' LCA(6 4));
Sortida
LCA(67) : 1 LCA(64) : 2
Nota: Estem calculant prèviament tota la potència requerida de 2 i també precalculant tots els valors de registre necessaris per garantir una complexitat de temps constant per consulta. En cas contrari, si féssim càlculs de registre per a cada operació de consulta, la nostra complexitat de temps no hauria estat constant.
Complexitat temporal: El procés de conversió de LCA a RMQ el fa Euler Walk que porta O(n) temps.
El preprocessament de la taula escassa a RMQ pren temps O(nlogn) i respondre cada consulta és un procés de temps constant. Per tant, la complexitat temporal global és O(nlogn) - preprocessament i O(1) per a cada consulta.
Espai auxiliar: O(n+s)
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