Priekšnosacījumi: BIT DFS
Ņemot vērā sakņotu koku T ar 'n' mezgliem, katram mezglam ir krāsa, ko apzīmē ar masīva krāsu [] (krāsa[i] apzīmē itā mezgla krāsu vesela skaitļa formā). Atbildiet uz šāda veida “Q” vaicājumiem:
- atšķiras u — izdrukājiet atšķirīgu krāsainu mezglu skaitu zem apakškoka, kas sakņojas zem burta "u"
Piemēri:
1
/
2 3
/| |
4 5 6 7 8
/|
9 10 11
color[] = {0 2 3 3 4 1 3 4 3 2 1 1}
Indexes NA 1 2 3 4 5 6 7 8 9 10 11
(Node Values and colors start from index 1)
distinct 3 -> output should be '4'.
There are six different nodes in the subtree rooted with
3 nodes are 3 7 8 9 10 and 11. These nodes have
four distinct colors (3 4 2 and 1)
distinct 2 -> output should be '3'.
distinct 7 -> output should be '3'.
Risinājuma izveide soļos:
- Izlīdziniet koku, izmantojot DFS; glabājiet katra mezgla apmeklējuma laiku un beigu laiku divos masīvos vis_time[i] saglabā i-tā mezgla apmeklējuma laiku beigu_laiks[i] saglabā beigu laiku.
- Tajā pašā DFS izsaukumā saglabājiet katra masīva mezgla krāsas vērtību plakans_koks[] pie indeksiem: vis_time[i] un beigu_laiks[i] i-tam mezglam.
Piezīme: masīva lielums plakans_koks[] būs 2n.
Tagad problēma ir samazināta līdz atšķirīgu elementu skaita atrašanai diapazonā [vis_time[u] beigu_laiks[u] ] masīvā plakans_koks[] katram norādītā veida vaicājumam. Lai to izdarītu, mēs apstrādāsim vaicājumus bezsaistē (apstrādājot vaicājumus tādā secībā, kas atšķiras no jautājumā norādītā, un saglabājot rezultātus un visbeidzot izdrukājot rezultātus katram jautājumā norādītajā secībā).
Darbības:
- Vispirms mēs iepriekš apstrādājam masīvu plakans_koks[] ; mēs uzturam a tabula[] (vektoru masīvs) galds [i] saglabā vektoru, kurā ir visi indeksi plakans_koks[] kam ir vērtība i. Tas ir, ja plakans_koks[j] = i tad galds [i] būs viens no tā elementiem j .
- BIT mēs atjauninām '1' i-tajā indeksā, ja vēlamies i-to elementu plakans_koks[] jāieskaita vaicājums() metodi. Tagad mēs uzturam citu masīvu šķērsot[] ; krusts [i] satur rādītāju uz nākamo tabulas [i] elementu, kas vēl nav atzīmēts BIT.
- Tagad mēs atjauninām savu BIT un iestatām “1” katram elementam pirmo reizi plakans_koks[] un atbilstošs pieaugums šķērsot[] ar '1' (ja plakans_koks[i] tad notiek pirmo reizi krusts[plakans_koks[i]] tiek palielināts par "1"), lai norādītu uz nākamo šī elementa atkārtojumu.
- Tagad mūsu vaicājums (R) funkcija BIT atgrieztu atšķirīgo elementu skaitu plakans_koks[] diapazonā [1 R] .
- Mēs sakārtojam visus vaicājumus pieauguma secībā izrādes_laiks[] ļaut l i apzīmēt vis_time[i] un r i apzīmē beigu_laiks[i] . Vaicājumu kārtošana augošā secībā l i dod mums priekšrocības, jo, apstrādājot i-to vaicājumu, mēs turpmāk neredzēsim nevienu vaicājumu ar tā " l ' mazāks par l i . Tātad mēs varam noņemt visus elementu notikumus līdz l i - 1 no BIT un pievienojiet to nākamos gadījumus, izmantojot šķērsot[] masīvs. Un tad vaicājums (R) atgrieztu atšķirīgo elementu skaitu diapazonā [l i r i ] .
// A C++ program implementing the above design #include #define max_color 1000005 #define maxn 100005 using namespace std; // Note: All elements of global arrays are // initially zero // All the arrays have been described above int bit[maxn] vis_time[maxn] end_time[maxn]; int flat_tree[2 * maxn]; vector<int> tree[maxn]; vector<int> table[max_color]; int traverser[max_color]; bool vis[maxn]; int tim = 0; //li ri and index are stored in queries vector //in that order as the sort function will use //the value li for comparison vector< pair< pair<int int> int> > queries; //ans[i] stores answer to ith query int ans[maxn]; //update function to add val to idx in BIT void update(int idx int val) { while ( idx < maxn ) { bit[idx] += val; idx += idx & -idx; } } //query function to find sum(1 idx) in BIT int query(int idx) { int res = 0; while ( idx > 0 ) { res += bit[idx]; idx -= idx & -idx; } return res; } void dfs(int v int color[]) { //mark the node visited vis[v] = 1; //set visiting time of the node v vis_time[v] = ++tim; //use the color of node v to fill flat_tree[] flat_tree[tim] = color[v]; vector<int>::iterator it; for (it=tree[v].begin(); it!=tree[v].end(); it++) if (!vis[*it]) dfs(*it color); // set ending time for node v end_time[v] = ++tim; // setting its color in flat_tree[] again flat_tree[tim] = color[v]; } //function to add an edge(u v) to the tree void addEdge(int u int v) { tree[u].push_back(v); tree[v].push_back(u); } //function to build the table[] and also add //first occurrences of elements to the BIT void hashMarkFirstOccurrences(int n) { for (int i = 1 ; i <= 2 * n ; i++) { table[flat_tree[i]].push_back(i); //if it is the first occurrence of the element //then add it to the BIT and increment traverser if (table[flat_tree[i]].size() == 1) { //add the occurrence to bit update(i 1); //make traverser point to next occurrence traverser[flat_tree[i]]++; } } } //function to process all the queries and store their answers void processQueries() { int j = 1; for (int i=0; i<queries.size(); i++) { //for each query remove all the occurrences before its li //li is the visiting time of the node //which is stored in first element of first pair for ( ; j < queries[i].first.first ; j++ ) { int elem = flat_tree[j]; //update(i -1) removes an element at ith index //in the BIT update( table[elem][traverser[elem] - 1] -1); //if there is another occurrence of the same element if ( traverser[elem] < table[elem].size() ) { //add the occurrence to the BIT and //increment traverser update(table[elem][ traverser[elem] ] 1); traverser[elem]++; } } //store the answer for the query the index of the query //is the second element of the pair //And ri is stored in second element of the first pair ans[queries[i].second] = query(queries[i].first.second); } } // Count distinct colors in subtrees rooted with qVer[0] // qVer[1] ...qVer[qn-1] void countDistinctColors(int color[] int n int qVer[] int qn) { // build the flat_tree[] vis_time[] and end_time[] dfs(1 color); // add query for u = 3 2 and 7 for (int i=0; i<qn; i++) queries.push_back(make_pair(make_pair(vis_time[qVer[i]] end_time[qVer[i]]) i) ); // sort the queries in order of increasing vis_time sort(queries.begin() queries.end()); // make table[] and set '1' at first occurrences of elements hashMarkFirstOccurrences(n); // process queries processQueries(); // print all the answers in order asked // in the question for (int i=0; i<queries.size() ; i++) { cout << 'Distinct colors in the corresponding subtree' 'is: ' << ans[i] << endl; } } //driver code int main() { /* 1 / 2 3 /| | 4 5 6 7 8 /| 9 10 11 */ int n = 11; int color[] = {0 2 3 3 4 1 3 4 3 2 1 1}; // add all the edges to the tree addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); addEdge(7 9); addEdge(7 10); addEdge(7 11); int qVer[] = {3 2 7}; int qn = sizeof(qVer)/sizeof(qVer[0]); countDistinctColors(color n qVer qn); return 0; }
Java import java.util.ArrayList; import java.util.Collections; import java.util.List; public class Main { private static final int maxColor = 1000005; private static final int maxn = 100005; private static int[] bit = new int[maxn]; private static int[] visTime = new int[maxn]; private static int[] endTime = new int[maxn]; private static int[] flatTree = new int[2 * maxn]; private static List<Integer>[] tree = new ArrayList[maxn]; private static List<Integer>[] table = new ArrayList[maxColor]; private static int[] traverser = new int[maxColor]; private static boolean[] vis = new boolean[maxn]; private static int tim = 0; private static List<Pair<Pair<Integer Integer> Integer>> queries = new ArrayList<>(); private static int[] ans = new int[maxn]; private static void update(int idx int val) { while (idx < maxn) { bit[idx] += val; idx += idx & -idx; } } private static int query(int idx) { int res = 0; while (idx > 0) { res += bit[idx]; idx -= idx & -idx; } return res; } private static void dfs(int v int[] color) { vis[v] = true; visTime[v] = ++tim; flatTree[tim] = color[v]; for (int u : tree[v]) { if (!vis[u]) { dfs(u color); } } endTime[v] = ++tim; flatTree[tim] = color[v]; } private static void addEdge(int u int v) { tree[u].add(v); tree[v].add(u); } private static void hashMarkFirstOccurrences(int n) { for (int i = 1; i <= 2 * n; i++) { table[flatTree[i]].add(i); if (table[flatTree[i]].size() == 1) { update(i 1); traverser[flatTree[i]]++; } } } private static void processQueries() { int j = 1; for (Pair<Pair<Integer Integer> Integer> query : queries) { for (; j < query.first.first; j++) { int elem = flatTree[j]; update(table[elem].get(traverser[elem] - 1) -1); if (traverser[elem] < table[elem].size()) { update(table[elem].get(traverser[elem]) 1); traverser[elem]++; } } ans[query.second] = query(query.first.second); } } private static void countDistinctColors(int[] color int n int[] qVer int qn) { dfs(1 color); for (int i = 0; i < qn; i++) { queries.add(new Pair<>(new Pair<>(visTime[qVer[i]] endTime[qVer[i]]) i)); } Collections.sort(queries); hashMarkFirstOccurrences(n); processQueries(); for (int i = 0; i < queries.size(); i++) { System.out.println('Distinct colors in the corresponding subtree is: ' + ans[i]); } } public static void main(String[] args) { int n = 11; int[] color = {0 2 3 3 4 1 3 4 3 2 1 1}; for (int i = 0; i < maxn; i++) { tree[i] = new ArrayList<>(); table[i] = new ArrayList<>(); // Initialize the table array here } addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); addEdge(7 9); addEdge(7 10); addEdge(7 11); int[] qVer = {3 2 7}; int qn = qVer.length; countDistinctColors(color n qVer qn); } static class Pair<A B> implements Comparable<Pair<A B>> { A first; B second; public Pair(A first B second) { this.first = first; this.second = second; } @Override public int compareTo(Pair<A B> other) { if (this.first.equals(other.first)) { return ((Comparable<B>) this.second).compareTo(other.second); } else { return ((Comparable<A>) this.first).compareTo(other.first); } } } }
Python3 # All elements of global arrays are initially zero bit = [0] * 100005 # Binary Indexed Tree (BIT) vis_time = [0] * 100005 # Visiting time for nodes end_time = [0] * 100005 # Ending time for nodes flat_tree = [0] * (2 * 100005) # Flattened tree array tree = [[] for _ in range(100005)] # Tree adjacency list table = [[] for _ in range(1000005)] # Table to store occurrences of colors traverser = [0] * 1000005 # Keeps track of occurrences for each color vis = [False] * 100005 # Visited nodes tim = 0 # Time variable for node traversal queries = [] # Queries to process ans = [0] * 100005 # Stores answers to queries # Update function to add val to idx in BIT def update(idx val): while idx < len(bit): bit[idx] += val idx += idx & -idx # Query function to find sum(1 idx) in BIT def query(idx): res = 0 while idx > 0: res += bit[idx] idx -= idx & -idx return res def dfs(v color): global tim vis[v] = True vis_time[v] = tim = tim + 1 flat_tree[tim] = color[v] # Flattening the tree with node colors for node in tree[v]: # Traverse through adjacent nodes if not vis[node]: dfs(node color) end_time[v] = tim = tim + 1 flat_tree[tim] = color[v] def addEdge(u v): tree[u].append(v) # Add edges to the tree tree[v].append(u) def hashMarkFirstOccurrences(n): # Loop through the flattened tree to mark first occurrences of colors for i in range(1 2 * n + 1): table[flat_tree[i]].append(i) if len(table[flat_tree[i]]) == 1: update(i 1) # Update BIT for first occurrences traverser[flat_tree[i]] += 1 def processQueries(): j = 1 for i in range(len(queries)): # Process queries and update BIT accordingly while j < queries[i][0][0]: elem = flat_tree[j] update(table[elem][traverser[elem] - 1] -1) if traverser[elem] < len(table[elem]): update(table[elem][traverser[elem]] 1) traverser[elem] += 1 j += 1 ans[queries[i][1]] = query(queries[i][0][1]) # Store query answers def countDistinctColors(color n qVer qn): dfs(1 color) # Start depth-first search from node 1 for i in range(qn): queries.append(((vis_time[qVer[i]] end_time[qVer[i]]) i)) # Prepare queries queries.sort() # Sort queries based on visiting time and ending time hashMarkFirstOccurrences(n) # Mark first occurrences in the flattened tree processQueries() # Process queries and update BIT for i in range(len(queries)): print(f'Distinct colors in the corresponding subtree is: {ans[i]}') # Print query answers if __name__ == '__main__': # Sample tree structure and colors n = 11 color = [0 2 3 3 4 1 3 4 3 2 1 1] addEdge(1 2) # Add edges to construct the tree addEdge(1 3) addEdge(2 4) addEdge(2 5) addEdge(2 6) addEdge(3 7) addEdge(3 8) addEdge(7 9) addEdge(7 10) addEdge(7 11) qVer = [3 2 7] # Query nodes qn = len(qVer) countDistinctColors(color n qVer qn) # Count distinct colors in subtrees rooted at query nodes
C# using System; using System.Collections.Generic; using System.Linq; class Program { // Note: All elements of global arrays are // initially zero // All the arrays have been described above const int max_color = 1000005; const int maxn = 100005; static int[] bit = new int[maxn]; static int[] vis_time = new int[maxn]; static int[] end_time = new int[maxn]; static int[] flat_tree = new int[2 * maxn]; static List<int>[] tree = Enumerable.Repeat(0 maxn).Select(x => new List<int>()).ToArray(); static List<int>[] table = Enumerable.Repeat(0 max_color).Select(x => new List<int>()).ToArray(); static int[] traverser = new int[max_color]; static bool[] vis = new bool[maxn]; static int tim = 0; // li ri and index are stored in queries vector // in that order as the sort function will use // the value li for comparison static List<Tuple<Tuple<int int> int>> queries = new List<Tuple<Tuple<int int> int>>(); // ans[i] stores answer to ith query static int[] ans = new int[maxn]; // update function to add val to idx in BIT static void Update(int idx int val) { while (idx < maxn) { bit[idx] += val; idx += idx & -idx; } } // query function to find sum(1 idx) in BIT static int Query(int idx) { int res = 0; while (idx > 0) { res += bit[idx]; idx -= idx & -idx; } return res; } static void Dfs(int v int[] color) { // mark the node visited vis[v] = true; vis_time[v] = ++tim; flat_tree[tim] = color[v]; foreach (int it in tree[v]) if (!vis[it]) Dfs(it color); end_time[v] = ++tim; flat_tree[tim] = color[v]; } //function to add edges to graph static void addEdge(int u int v) { tree[u].Add(v); tree[v].Add(u); } // function to build the table[] and also add // first occurrences of elements to the BIT static void HashMarkFirstOccurrences(int n) { for (int i = 1; i <= 2 * n; i++) { // if it is the first occurrence of the element // then add it to the BIT and increment traverser table[flat_tree[i]].Add(i); if (table[flat_tree[i]].Count == 1) { Update(i 1); traverser[flat_tree[i]]++; } } } // function to process all the queries and store their answers static void ProcessQueries() { int j = 1; // for each query remove all the occurrences before its li // li is the visiting time of the node // which is stored in first element of first pair for (int i = 0; i < queries.Count; i++) { for (; j < queries[i].Item1.Item1; j++) { int elem = flat_tree[j]; Update(table[elem][traverser[elem] - 1] -1); if (traverser[elem] < table[elem].Count) { Update(table[elem][traverser[elem]] 1); traverser[elem]++; } } ans[queries[i].Item2] = Query(queries[i].Item1.Item2); } } // Count distinct colors in subtrees rooted with qVer[0] // qVer[1] ...qVer[qn-1] static void countDistinctColors(int[] color int n int[] qVer int qn) { // build the flat_tree[] vis_time[] and end_time[] Dfs(1 color); // add query for u = 3 2 and 7 for (int i = 0; i < qn; i++) queries.Add(new Tuple<Tuple<int int> int>(new Tuple<int int>(vis_time[qVer[i]] end_time[qVer[i]]) i)); queries.Sort(); HashMarkFirstOccurrences(n); ProcessQueries(); // print all the answers in order asked // in the question for (int i = 0; i < queries.Count; i++) Console.WriteLine('Distinct colors in the corresponding subtree is: {0}' ans[i]); } static void Main(string[] args) { /* 1 / 2 3 /| | 4 5 6 7 8 /| 9 10 11 */ int n = 11; int[] color = { 0 2 3 3 4 1 3 4 3 2 1 1 }; // add all the edges to the tree addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); addEdge(7 9); addEdge(7 10); addEdge(7 11); int[] qVer = { 3 2 7 }; int qn = qVer.Length; countDistinctColors(color n qVer qn); } }
JavaScript // Constants for maximum color maximum nodes and initializing arrays const max_color = 1000005; const maxn = 100005; const bit = new Array(maxn).fill(0); // Binary Indexed Tree const vis_time = new Array(maxn).fill(0); // Visit time for nodes const end_time = new Array(maxn).fill(0); // End time for nodes const flat_tree = new Array(2 * maxn).fill(0); // Flattened tree structure const tree = Array.from({ length: maxn } () => []); // Graph/tree structure const table = Array.from({ length: max_color } () => []); // Table for elements' occurrences const traverser = new Array(max_color).fill(0); // Tracks traversed elements const vis = new Array(maxn).fill(false); // Tracks visited nodes let tim = 0; // Time counter const queries = []; // Array to store queries const ans = new Array(maxn).fill(0); // Array to store answers to queries // Function to update Binary Indexed Tree function Update(idx val) { while (idx < maxn) { bit[idx] += val; idx += idx & -idx; } } // Function to query Binary Indexed Tree function Query(idx) { let res = 0; while (idx > 0) { res += bit[idx]; idx -= idx & -idx; } return res; } // Depth-first search traversal on the tree function Dfs(v color) { vis[v] = true; vis_time[v] = ++tim; flat_tree[tim] = color[v]; tree[v].forEach((it) => { if (!vis[it]) Dfs(it color); }); end_time[v] = ++tim; flat_tree[tim] = color[v]; } // Function to add edges to the tree/graph function addEdge(u v) { tree[u].push(v); tree[v].push(u); } // Function to populate table and BIT with first occurrences function HashMarkFirstOccurrences(n) { for (let i = 1; i <= 2 * n; i++) { table[flat_tree[i]].push(i); if (table[flat_tree[i]].length === 1) { Update(i 1); traverser[flat_tree[i]]++; } } } // Function to process queries and store answers function ProcessQueries() { let j = 1; for (let i = 0; i < queries.length; i++) { for (; j < queries[i][0][0]; j++) { const elem = flat_tree[j]; Update(table[elem][traverser[elem] - 1] -1); if (traverser[elem] < table[elem].length) { Update(table[elem][traverser[elem]] 1); traverser[elem]++; } } ans[queries[i][1]] = Query(queries[i][0][1]); } } // Function to count distinct colors in subtrees function countDistinctColors(color n qVer qn) { Dfs(1 color); // Traverse the tree to generate visit and end times for (let i = 0; i < qn; i++) { // Push queries based on visit and end times to queries array queries.push([[vis_time[qVer[i]] end_time[qVer[i]]] i]); } queries.sort((a b) => a[0][0] - b[0][0]); // Sort queries based on visit times HashMarkFirstOccurrences(n); // Initialize BIT and table with first occurrences ProcessQueries(); // Process queries to calculate distinct colors for (let i = 0; i < queries.length; i++) { console.log(`Distinct colors in the corresponding subtree is: ${ans[i]}`); // Print the answers } } // Define the tree structure and colors const n = 11; const color = [0 2 3 3 4 1 3 4 3 2 1 1]; // Define edges in the tree addEdge(1 2); addEdge(1 3); addEdge(2 4); addEdge(2 5); addEdge(2 6); addEdge(3 7); addEdge(3 8); addEdge(7 9); addEdge(7 10); addEdge(7 11); // Define query vertices and call the function to count distinct colors const qVer = [3 2 7]; const qn = qVer.length; countDistinctColors(color n qVer qn);
Izvade:
Distinct colors in the corresponding subtree is:4
Distinct colors in the corresponding subtree is:3
Distinct colors in the corresponding subtree is:3
Time Complexity: O( Q * log(n) )