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Binarno indeksirano drevo: posodobitev obsega in poizvedbe po obsegu

Podana je matrika arr[0..N-1]. Izvesti je treba naslednje operacije. 

  1. posodobitev (l r val) : Dodajte 'val' vsem elementom v matriki od [l r].
  2. getRangeSum(l r) : Poiščite vsoto vseh elementov v matriki iz [l r].

Na začetku so vsi elementi v matriki 0. Poizvedbe so lahko v poljubnem vrstnem redu, tj. pred vsoto obsega je lahko veliko posodobitev.



primer:

Vnos: N = 5 // {0 0 0 0 0}
Poizvedbe: posodobitev: l = 0 r = 4 val = 2
               posodobitev: l = 3 r = 4 val = 3 
               getRangeSum : l = 2 r = 4

Izhod: Vsota elementov obsega [2 4] je 12
Pojasnilo: Matrika po prvi posodobitvi postane {2 2 2 2 2}
Matrika po drugi posodobitvi postane {2 2 2 5 5}



Naivni pristop: Za rešitev težave sledite spodnji zamisli:

V prejšnja objava razpravljali smo o posodobitvi obsega in rešitvah za poizvedovanje točk z uporabo BIT. 
rangeUpdate(l r val) : dodamo 'val' elementu pri indeksu 'l'. Odštejemo 'val' od elementa z indeksom 'r+1'. 
getElement(index) [ali getSum()]: vrnemo vsoto elementov od 0 do indeksa, ki ga lahko hitro pridobimo z BIT.
RangSum() lahko izračunamo s poizvedbami getSum(). 
obsegSum(l r) = getSum(r) - getSum(l-1)

koliko mest je v Združenih državah

Preprosta rešitev je uporaba rešitev, obravnavanih v prejšnja objava . Poizvedba za posodobitev obsega je enaka. Poizvedbo za vsoto obsega lahko dosežete tako, da izvedete poizvedbo za pridobitev za vse elemente v obsegu. 



Učinkovit pristop: Za rešitev težave sledite spodnji zamisli:

Vsoto obsega dobimo s pomočjo vsot predpon. Kako zagotoviti, da je posodobitev izvedena tako, da je vsoto predpone mogoče izvesti hitro? Razmislite o situaciji, kjer je vsota predpone [0 k] (kjer je 0<= k < n) is needed after range update on the range [l r]. Three cases arise as k can possibly lie in 3 regions.

  • Primer 1 : 0< k < l 
    • Poizvedba za posodobitev ne bo vplivala na poizvedbo vsota.
  • Primer 2 : l<= k <= r 
    • Razmislite o primeru:  Dodajte 2 obsegu [2 4], rezultat pa bi bil: 0 0 2 2 2
      Če je k = 3 Vsota iz [0 k] = 4

Kako do tega rezultata? 
Preprosto dodajte val iz lthkazalo k kthkazalo. Vsota se po poizvedbi za posodobitev poveča za 'val*(k) - val*(l-1)'. 

  • Primer 3 : k > r 
    • V tem primeru moramo dodati 'val' iz lthkazalo na rthkazalo. Vsota se poveča za 'val*r – val*(l-1)' zaradi poizvedbe za posodobitev.

Opažanja:  

1. primer: je preprosta, saj bi vsota ostala enaka, kot je bila pred posodobitvijo.

java podniz vsebuje

2. primer: Vsota je bila povečana za val*k - val*(l-1). Najdemo lahko 'val', to je podobno iskanju ithelement v posodobitev obsega in članek o poizvedbi točk . Zato vzdržujemo en BIT za posodobitev obsega in poizvedbe točk, ta BIT bo v pomoč pri iskanju vrednosti pri kthkazalo. Zdaj je val * k izračunan, kako ravnati z dodatnim izrazom val*(l-1)? 
Za obravnavo tega dodatnega roka vzdržujemo še en BIT (BIT2). Posodobi vrednost * (l-1) pri lthindeks, tako da bo, ko se izvede poizvedba getSum na BIT2, dal rezultat kot val*(l-1).

Primer 3: Vsota v primeru 3 je bila povečana za 'val*r - val *(l-1)' vrednost tega izraza je mogoče pridobiti z uporabo BIT2. Namesto seštevanja odštejemo 'val*(l-1) - val*r', saj lahko dobimo to vrednost iz BIT2 z dodajanjem val*(l-1), kot smo storili v primeru 2, in odštevanjem val*r pri vsaki operaciji posodabljanja.

Posodobi poizvedbo 

Posodobitev (BITree1 l val)
Posodobitev (BITree1 r+1 -val)
PosodobitevBIT2(BITree2 l val*(l-1))
UpdateBIT2(BITree2 r+1 -val*r)

obseg vsota 

getSum(BITTree1 k) *k) - getSum(BITTree2 k)

Za rešitev težave sledite spodnjim korakom:

  • Ustvarite dve binarni indeksni drevesi z uporabo dane funkcije constructBITree()
  • Če želite najti vsoto v danem obsegu, pokličite funkcijo rangeSum() s parametri kot dani obseg in binarno indeksirana drevesa
    • Pokličite funkcijo vsota, ki bo vrnila vsoto v obsegu [0 X]
    • Vrni vsoto(R) - vsoto(L-1)
      • Znotraj te funkcije pokličite funkcijo getSum(), ki bo vrnila vsoto matrike iz [0 X]
      • Vrni getSum(Tree1 x) * x - getSum(tree2 x)
      • Znotraj funkcije getSum() ustvarite celoštevilsko vsoto enako nič in povečajte indeks za 1
      • Medtem ko je indeks večji od nič, povečajte vsoto za Tree[index]
      • Zmanjšajte indeks za (index & (-index)), da premaknete indeks na nadrejeno vozlišče v drevesu
      • Povratna vsota
  • Izpišite vsoto v danem območju

Spodaj je izvedba zgornjega pristopa: 

C++ 
// C++ program to demonstrate Range Update // and Range Queries using BIT #include    using namespace std; // Returns sum of arr[0..index]. This function assumes // that the array is preprocessed and partial sums of // array elements are stored in BITree[] int getSum(int BITree[] int index) {  int sum = 0; // Initialize result  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse ancestors of BITree[index]  while (index > 0) {  // Add current element of BITree to sum  sum += BITree[index];  // Move index to parent node in getSum View  index -= index & (-index);  }  return sum; } // Updates a node in Binary Index Tree (BITree) at given // index in BITree. The given value 'val' is added to // BITree[i] and all of its ancestors in tree. void updateBIT(int BITree[] int n int index int val) {  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;  // Traverse all ancestors and add 'val'  while (index <= n) {  // Add 'val' to current node of BI Tree  BITree[index] += val;  // Update index to that of parent in update View  index += index & (-index);  } } // Returns the sum of array from [0 x] int sum(int x int BITTree1[] int BITTree2[]) {  return (getSum(BITTree1 x) * x) - getSum(BITTree2 x); } void updateRange(int BITTree1[] int BITTree2[] int n  int val int l int r) {  // Update Both the Binary Index Trees  // As discussed in the article  // Update BIT1  updateBIT(BITTree1 n l val);  updateBIT(BITTree1 n r + 1 -val);  // Update BIT2  updateBIT(BITTree2 n l val * (l - 1));  updateBIT(BITTree2 n r + 1 -val * r); } int rangeSum(int l int r int BITTree1[] int BITTree2[]) {  // Find sum from [0r] then subtract sum  // from [0l-1] in order to find sum from  // [lr]  return sum(r BITTree1 BITTree2)  - sum(l - 1 BITTree1 BITTree2); } int* constructBITree(int n) {  // Create and initialize BITree[] as 0  int* BITree = new int[n + 1];  for (int i = 1; i <= n; i++)  BITree[i] = 0;  return BITree; } // Driver code int main() {  int n = 5;  // Construct two BIT  int *BITTree1 *BITTree2;  // BIT1 to get element at any index  // in the array  BITTree1 = constructBITree(n);  // BIT 2 maintains the extra term  // which needs to be subtracted  BITTree2 = constructBITree(n);  // Add 5 to all the elements from [04]  int l = 0 r = 4 val = 5;  updateRange(BITTree1 BITTree2 n val l r);  // Add 10 to all the elements from [24]  l = 2 r = 4 val = 10;  updateRange(BITTree1 BITTree2 n val l r);  // Find sum of all the elements from  // [14]  l = 1 r = 4;  cout << 'Sum of elements from [' << l << '' << r  << '] is ';  cout << rangeSum(l r BITTree1 BITTree2) << 'n';  return 0; }
Java 
// Java program to demonstrate Range Update // and Range Queries using BIT import java.util.*; class GFG {  // Returns sum of arr[0..index]. This function assumes  // that the array is preprocessed and partial sums of  // array elements are stored in BITree[]  static int getSum(int BITree[] int index)  {  int sum = 0; // Initialize result  // index in BITree[] is 1 more than the index in  // arr[]  index = index + 1;  // Traverse ancestors of BITree[index]  while (index > 0) {  // Add current element of BITree to sum  sum += BITree[index];  // Move index to parent node in getSum View  index -= index & (-index);  }  return sum;  }  // Updates a node in Binary Index Tree (BITree) at given  // index in BITree. The given value 'val' is added to  // BITree[i] and all of its ancestors in tree.  static void updateBIT(int BITree[] int n int index  int val)  {  // index in BITree[] is 1 more than the index in  // arr[]  index = index + 1;  // Traverse all ancestors and add 'val'  while (index <= n) {  // Add 'val' to current node of BI Tree  BITree[index] += val;  // Update index to that of parent in update View  index += index & (-index);  }  }  // Returns the sum of array from [0 x]  static int sum(int x int BITTree1[] int BITTree2[])  {  return (getSum(BITTree1 x) * x)  - getSum(BITTree2 x);  }  static void updateRange(int BITTree1[] int BITTree2[]  int n int val int l int r)  {  // Update Both the Binary Index Trees  // As discussed in the article  // Update BIT1  updateBIT(BITTree1 n l val);  updateBIT(BITTree1 n r + 1 -val);  // Update BIT2  updateBIT(BITTree2 n l val * (l - 1));  updateBIT(BITTree2 n r + 1 -val * r);  }  static int rangeSum(int l int r int BITTree1[]  int BITTree2[])  {  // Find sum from [0r] then subtract sum  // from [0l-1] in order to find sum from  // [lr]  return sum(r BITTree1 BITTree2)  - sum(l - 1 BITTree1 BITTree2);  }  static int[] constructBITree(int n)  {  // Create and initialize BITree[] as 0  int[] BITree = new int[n + 1];  for (int i = 1; i <= n; i++)  BITree[i] = 0;  return BITree;  }  // Driver Program to test above function  public static void main(String[] args)  {  int n = 5;  // Contwo BIT  int[] BITTree1;  int[] BITTree2;  // BIT1 to get element at any index  // in the array  BITTree1 = constructBITree(n);  // BIT 2 maintains the extra term  // which needs to be subtracted  BITTree2 = constructBITree(n);  // Add 5 to all the elements from [04]  int l = 0 r = 4 val = 5;  updateRange(BITTree1 BITTree2 n val l r);  // Add 10 to all the elements from [24]  l = 2;  r = 4;  val = 10;  updateRange(BITTree1 BITTree2 n val l r);  // Find sum of all the elements from  // [14]  l = 1;  r = 4;  System.out.print('Sum of elements from [' + l + ''  + r + '] is ');  System.out.print(rangeSum(l r BITTree1 BITTree2)  + 'n');  } } // This code is contributed by 29AjayKumar
Python3 
# Python3 program to demonstrate Range Update # and Range Queries using BIT # Returns sum of arr[0..index]. This function assumes # that the array is preprocessed and partial sums of # array elements are stored in BITree[] def getSum(BITree: list index: int) -> int: summ = 0 # Initialize result # index in BITree[] is 1 more than the index in arr[] index = index + 1 # Traverse ancestors of BITree[index] while index > 0: # Add current element of BITree to sum summ += BITree[index] # Move index to parent node in getSum View index -= index & (-index) return summ # Updates a node in Binary Index Tree (BITree) at given # index in BITree. The given value 'val' is added to # BITree[i] and all of its ancestors in tree. def updateBit(BITTree: list n: int index: int val: int) -> None: # index in BITree[] is 1 more than the index in arr[] index = index + 1 # Traverse all ancestors and add 'val' while index <= n: # Add 'val' to current node of BI Tree BITTree[index] += val # Update index to that of parent in update View index += index & (-index) # Returns the sum of array from [0 x] def summation(x: int BITTree1: list BITTree2: list) -> int: return (getSum(BITTree1 x) * x) - getSum(BITTree2 x) def updateRange(BITTree1: list BITTree2: list n: int val: int l: int r: int) -> None: # Update Both the Binary Index Trees # As discussed in the article # Update BIT1 updateBit(BITTree1 n l val) updateBit(BITTree1 n r + 1 -val) # Update BIT2 updateBit(BITTree2 n l val * (l - 1)) updateBit(BITTree2 n r + 1 -val * r) def rangeSum(l: int r: int BITTree1: list BITTree2: list) -> int: # Find sum from [0r] then subtract sum # from [0l-1] in order to find sum from # [lr] return summation(r BITTree1 BITTree2) - summation( l - 1 BITTree1 BITTree2) # Driver Code if __name__ == '__main__': n = 5 # BIT1 to get element at any index # in the array BITTree1 = [0] * (n + 1) # BIT 2 maintains the extra term # which needs to be subtracted BITTree2 = [0] * (n + 1) # Add 5 to all the elements from [04] l = 0 r = 4 val = 5 updateRange(BITTree1 BITTree2 n val l r) # Add 10 to all the elements from [24] l = 2 r = 4 val = 10 updateRange(BITTree1 BITTree2 n val l r) # Find sum of all the elements from # [14] l = 1 r = 4 print('Sum of elements from [%d%d] is %d' % (l r rangeSum(l r BITTree1 BITTree2))) # This code is contributed by # sanjeev2552
C# 
// C# program to demonstrate Range Update // and Range Queries using BIT using System; class GFG {  // Returns sum of arr[0..index]. This function assumes  // that the array is preprocessed and partial sums of  // array elements are stored in BITree[]  static int getSum(int[] BITree int index)  {  int sum = 0; // Initialize result  // index in BITree[] is 1 more than  // the index in []arr  index = index + 1;  // Traverse ancestors of BITree[index]  while (index > 0) {  // Add current element of BITree to sum  sum += BITree[index];  // Move index to parent node in getSum View  index -= index & (-index);  }  return sum;  }  // Updates a node in Binary Index Tree (BITree) at given  // index in BITree. The given value 'val' is added to  // BITree[i] and all of its ancestors in tree.  static void updateBIT(int[] BITree int n int index  int val)  {  // index in BITree[] is 1 more than  // the index in []arr  index = index + 1;  // Traverse all ancestors and add 'val'  while (index <= n) {  // Add 'val' to current node of BI Tree  BITree[index] += val;  // Update index to that of  // parent in update View  index += index & (-index);  }  }  // Returns the sum of array from [0 x]  static int sum(int x int[] BITTree1 int[] BITTree2)  {  return (getSum(BITTree1 x) * x)  - getSum(BITTree2 x);  }  static void updateRange(int[] BITTree1 int[] BITTree2  int n int val int l int r)  {  // Update Both the Binary Index Trees  // As discussed in the article  // Update BIT1  updateBIT(BITTree1 n l val);  updateBIT(BITTree1 n r + 1 -val);  // Update BIT2  updateBIT(BITTree2 n l val * (l - 1));  updateBIT(BITTree2 n r + 1 -val * r);  }  static int rangeSum(int l int r int[] BITTree1  int[] BITTree2)  {  // Find sum from [0r] then subtract sum  // from [0l-1] in order to find sum from  // [lr]  return sum(r BITTree1 BITTree2)  - sum(l - 1 BITTree1 BITTree2);  }  static int[] constructBITree(int n)  {  // Create and initialize BITree[] as 0  int[] BITree = new int[n + 1];  for (int i = 1; i <= n; i++)  BITree[i] = 0;  return BITree;  }  // Driver Code  public static void Main(String[] args)  {  int n = 5;  // Contwo BIT  int[] BITTree1;  int[] BITTree2;  // BIT1 to get element at any index  // in the array  BITTree1 = constructBITree(n);  // BIT 2 maintains the extra term  // which needs to be subtracted  BITTree2 = constructBITree(n);  // Add 5 to all the elements from [04]  int l = 0 r = 4 val = 5;  updateRange(BITTree1 BITTree2 n val l r);  // Add 10 to all the elements from [24]  l = 2;  r = 4;  val = 10;  updateRange(BITTree1 BITTree2 n val l r);  // Find sum of all the elements from  // [14]  l = 1;  r = 4;  Console.Write('Sum of elements from [' + l + '' + r  + '] is ');  Console.Write(rangeSum(l r BITTree1 BITTree2)  + 'n');  } } // This code is contributed by 29AjayKumar
JavaScript 
<script> // JavaScript program to demonstrate Range Update // and Range Queries using BIT // Returns sum of arr[0..index]. This function assumes // that the array is preprocessed and partial sums of // array elements are stored in BITree[] function getSum(BITreeindex) {  let sum = 0; // Initialize result    // index in BITree[] is 1 more than the index in arr[]  index = index + 1;    // Traverse ancestors of BITree[index]  while (index > 0)  {  // Add current element of BITree to sum  sum += BITree[index];    // Move index to parent node in getSum View  index -= index & (-index);  }  return sum; } // Updates a node in Binary Index Tree (BITree) at given // index in BITree. The given value 'val' is added to // BITree[i] and all of its ancestors in tree. function updateBIT(BITreenindexval) {  // index in BITree[] is 1 more than the index in arr[]  index = index + 1;    // Traverse all ancestors and add 'val'  while (index <= n)  {  // Add 'val' to current node of BI Tree  BITree[index] += val;    // Update index to that of parent in update View  index += index & (-index);  } } // Returns the sum of array from [0 x] function sum(xBITTree1BITTree2) {  return (getSum(BITTree1 x) * x) - getSum(BITTree2 x); } function updateRange(BITTree1BITTree2nvallr) {  // Update Both the Binary Index Trees  // As discussed in the article    // Update BIT1  updateBIT(BITTree1 n l val);  updateBIT(BITTree1 n r + 1 -val);    // Update BIT2  updateBIT(BITTree2 n l val * (l - 1));  updateBIT(BITTree2 n r + 1 -val * r); } function rangeSum(lrBITTree1BITTree2) {  // Find sum from [0r] then subtract sum  // from [0l-1] in order to find sum from  // [lr]  return sum(r BITTree1 BITTree2) -  sum(l - 1 BITTree1 BITTree2); } function constructBITree(n) {  // Create and initialize BITree[] as 0  let BITree = new Array(n + 1);  for (let i = 1; i <= n; i++)  BITree[i] = 0;    return BITree; } // Driver Program to test above function let n = 5;   // Contwo BIT let BITTree1; let BITTree2; // BIT1 to get element at any index // in the array BITTree1 = constructBITree(n); // BIT 2 maintains the extra term // which needs to be subtracted BITTree2 = constructBITree(n); // Add 5 to all the elements from [04] let l = 0  r = 4  val = 5; updateRange(BITTree1 BITTree2 n val l r); // Add 10 to all the elements from [24] l = 2 ; r = 4 ; val = 10; updateRange(BITTree1 BITTree2 n val l r); // Find sum of all the elements from // [14] l = 1 ; r = 4; document.write('Sum of elements from [' + l  + '' + r+ '] is '); document.write(rangeSum(l r BITTree1 BITTree2)+ '  
'); // This code is contributed by rag2127 </script>

Izhod 
Sum of elements from [14] is 50

Časovna zapletenost : O(q * log(N)), kjer je q število poizvedb.
Pomožni prostor: O(N)