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Divdimensiju binārs indeksēts koks vai fenvikas koks

Priekšnoteikums - Fenvikas koks

Mēs zinām, ka, lai atbildētu uz diapazona summas vaicājumiem 1D masīvā, efektīvi binārs indeksēts koks (vai Fenwick Tree) ir labākā izvēle (pat labāk nekā segmenta koks, jo ir mazāk atmiņas prasības un nedaudz ātrāk nekā segmenta koks).



Vai mēs varam efektīvi atbildēt uz apakšmatrikas summas jautājumiem, izmantojot bināro indeksēto koku?

Atbilde ir Apvidū Tas ir iespējams, izmantojot a 2D bits kas nav nekas cits kā 1D bitu masīvs. 

Algoritms:

Mēs apsveram zemāk esošo piemēru. Pieņemsim, ka mums jāatrod visu numuru summa, kas atrodas izceltajā apgabalā:



Fenvikas koks' title=

Mēs pieņemam, ka matricas izcelsme ir apakšā - O.Then 2D bits izmanto faktu, ka-

Sum under the marked area = Sum(OB) - Sum(OD) - Sum(OA) + Sum(OC) 

Divdimensiju binārs indeksēts koks vai fenvikas koks



Mūsu programmā mēs izmantojam GetSum (X Y) funkciju, kas atrod matricas summu no (0 0) līdz (x y). 
Tādējādi zemāk esošā formula: 

Sum under the marked area = Sum(OB) - Sum(OD) - Sum(OA) + Sum(OC) The above formula gets reduced to Query(x1y1x2y2) = getSum(x2 y2) - getSum(x2 y1-1) - getSum(x1-1 y2) + getSum(x1-1 y1-1) 

kur 
x1 y1 = x un y koordinātas c 
x2 y2 = x un y koordinātas B
Funkcija UpdateBit (x y val) atjaunina visus elementus zem reģiona - (x y) līdz (n m), kur 
N = Visas matricas maksimālais x koordināts. 
M = Visas matricas maksimālais y koordināts.
Atpūtas procedūra ir diezgan līdzīga 1D binārā indeksētā koka procedūrai.

Zemāk ir 2D indeksētā koka C ++ ieviešana 

C++
/* C++ program to implement 2D Binary Indexed Tree 2D BIT is basically a BIT where each element is another BIT. Updating by adding v on (x y) means it's effect will be found throughout the rectangle [(x y) (max_x max_y)] and query for (x y) gives you the result of the rectangle [(0 0) (x y)] assuming the total rectangle is [(0 0) (max_x max_y)]. So when you query and update on this BITyou have to be careful about how many times you are subtracting a rectangle and adding it. Simple set union formula works here. So if you want to get the result of a specific rectangle [(x1 y1) (x2 y2)] the following steps are necessary: Query(x1y1x2y2) = getSum(x2 y2)-getSum(x2 y1-1) -  getSum(x1-1 y2)+getSum(x1-1 y1-1) Here 'Query(x1y1x2y2)' means the sum of elements enclosed in the rectangle with bottom-left corner's co-ordinates (x1 y1) and top-right corner's co-ordinates - (x2 y2) Constraints -> x1<=x2 and y1<=y2  /  y |  | --------(x2y2)  | | |  | | |  | | |  | ---------  | (x1y1)  |  |___________________________  (0 0) x--> In this program we have assumed a square matrix. The program can be easily extended to a rectangular one. */ #include   using namespace std; #define N 4 // N-->max_x and max_y // A structure to hold the queries struct Query {  int x1 y1; // x and y co-ordinates of bottom left  int x2 y2; // x and y co-ordinates of top right }; // A function to update the 2D BIT void updateBIT(int BIT[][N+1] int x int y int val) {  for (; x <= N; x += (x & -x))  {  // This loop update all the 1D BIT inside the  // array of 1D BIT = BIT[x]  for (int yy=y; yy <= N; yy += (yy & -yy))  BIT[x][yy] += val;  }  return; } // A function to get sum from (0 0) to (x y) int getSum(int BIT[][N+1] int x int y) {  int sum = 0;  for(; x > 0; x -= x&-x)  {  // This loop sum through all the 1D BIT  // inside the array of 1D BIT = BIT[x]  for(int yy=y; yy > 0; yy -= yy&-yy)  {  sum += BIT[x][yy];  }  }  return sum; } // A function to create an auxiliary matrix // from the given input matrix void constructAux(int mat[][N] int aux[][N+1]) {  // Initialise Auxiliary array to 0  for (int i=0; i<=N; i++)  for (int j=0; j<=N; j++)  aux[i][j] = 0;  // Construct the Auxiliary Matrix  for (int j=1; j<=N; j++)  for (int i=1; i<=N; i++)  aux[i][j] = mat[N-j][i-1];  return; } // A function to construct a 2D BIT void construct2DBIT(int mat[][N] int BIT[][N+1]) {  // Create an auxiliary matrix  int aux[N+1][N+1];  constructAux(mat aux);  // Initialise the BIT to 0  for (int i=1; i<=N; i++)  for (int j=1; j<=N; j++)  BIT[i][j] = 0;  for (int j=1; j<=N; j++)  {  for (int i=1; i<=N; i++)  {  // Creating a 2D-BIT using update function  // everytime we/ encounter a value in the  // input 2D-array  int v1 = getSum(BIT i j);  int v2 = getSum(BIT i j-1);  int v3 = getSum(BIT i-1 j-1);  int v4 = getSum(BIT i-1 j);  // Assigning a value to a particular element  // of 2D BIT  updateBIT(BIT i j aux[i][j]-(v1-v2-v4+v3));  }  }  return; } // A function to answer the queries void answerQueries(Query q[] int m int BIT[][N+1]) {  for (int i=0; i<m; i++)  {  int x1 = q[i].x1 + 1;  int y1 = q[i].y1 + 1;  int x2 = q[i].x2 + 1;  int y2 = q[i].y2 + 1;  int ans = getSum(BIT x2 y2)-getSum(BIT x2 y1-1)-  getSum(BIT x1-1 y2)+getSum(BIT x1-1 y1-1);  printf ('Query(%d %d %d %d) = %dn'  q[i].x1 q[i].y1 q[i].x2 q[i].y2 ans);  }  return; } // Driver program int main() {  int mat[N][N] = {{1 2 3 4}  {5 3 8 1}  {4 6 7 5}  {2 4 8 9}};  // Create a 2D Binary Indexed Tree  int BIT[N+1][N+1];  construct2DBIT(mat BIT);  /* Queries of the form - x1 y1 x2 y2  For example the query- {1 1 3 2} means the sub-matrix-  y  /  3 | 1 2 3 4 Sub-matrix  2 | 5 3 8 1 {1132} ---> 3 8 1  1 | 4 6 7 5 6 7 5  0 | 2 4 8 9  |  --|------ 0 1 2 3 ----> x  |  Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30  */  Query q[] = {{1 1 3 2} {2 3 3 3} {1 1 1 1}};  int m = sizeof(q)/sizeof(q[0]);  answerQueries(q m BIT);  return(0); } 
Java
/* Java program to implement 2D Binary Indexed Tree  2D BIT is basically a BIT where each element is another BIT.  Updating by adding v on (x y) means it's effect will be found  throughout the rectangle [(x y) (max_x max_y)]  and query for (x y) gives you the result of the rectangle  [(0 0) (x y)] assuming the total rectangle is  [(0 0) (max_x max_y)]. So when you query and update on  this BITyou have to be careful about how many times you are  subtracting a rectangle and adding it. Simple set union formula  works here.  So if you want to get the result of a specific rectangle  [(x1 y1) (x2 y2)] the following steps are necessary:  Query(x1y1x2y2) = getSum(x2 y2)-getSum(x2 y1-1) -   getSum(x1-1 y2)+getSum(x1-1 y1-1)  Here 'Query(x1y1x2y2)' means the sum of elements enclosed  in the rectangle with bottom-left corner's co-ordinates  (x1 y1) and top-right corner's co-ordinates - (x2 y2)  Constraints -> x1<=x2 and y1<=y2   /  y |   | --------(x2y2)   | | |   | | |   | | |   | ---------   | (x1y1)   |   |___________________________  (0 0) x-->  In this program we have assumed a square matrix. The  program can be easily extended to a rectangular one. */ class GFG {  static final int N = 4; // N-.max_x and max_y  // A structure to hold the queries  static class Query  {   int x1 y1; // x and y co-ordinates of bottom left   int x2 y2; // x and y co-ordinates of top right   public Query(int x1 int y1 int x2 int y2)  {  this.x1 = x1;  this.y1 = y1;  this.x2 = x2;  this.y2 = y2;  }   };  // A function to update the 2D BIT  static void updateBIT(int BIT[][] int x  int y int val)  {   for (; x <= N; x += (x & -x))   {   // This loop update all the 1D BIT inside the   // array of 1D BIT = BIT[x]   for (; y <= N; y += (y & -y))   BIT[x][y] += val;   }   return;  }  // A function to get sum from (0 0) to (x y)  static int getSum(int BIT[][] int x int y)  {   int sum = 0;   for(; x > 0; x -= x&-x)   {   // This loop sum through all the 1D BIT   // inside the array of 1D BIT = BIT[x]   for(; y > 0; y -= y&-y)   {   sum += BIT[x][y];   }   }   return sum;  }  // A function to create an auxiliary matrix  // from the given input matrix  static void constructAux(int mat[][] int aux[][])  {   // Initialise Auxiliary array to 0   for (int i = 0; i <= N; i++)   for (int j = 0; j <= N; j++)   aux[i][j] = 0;   // Construct the Auxiliary Matrix   for (int j = 1; j <= N; j++)   for (int i = 1; i <= N; i++)   aux[i][j] = mat[N - j][i - 1];   return;  }  // A function to construct a 2D BIT  static void construct2DBIT(int mat[][]  int BIT[][])  {   // Create an auxiliary matrix   int [][]aux = new int[N + 1][N + 1];   constructAux(mat aux);   // Initialise the BIT to 0   for (int i = 1; i <= N; i++)   for (int j = 1; j <= N; j++)   BIT[i][j] = 0;   for (int j = 1; j <= N; j++)   {   for (int i = 1; i <= N; i++)   {   // Creating a 2D-BIT using update function   // everytime we/ encounter a value in the   // input 2D-array   int v1 = getSum(BIT i j);   int v2 = getSum(BIT i j - 1);   int v3 = getSum(BIT i - 1 j - 1);   int v4 = getSum(BIT i - 1 j);   // Assigning a value to a particular element   // of 2D BIT   updateBIT(BIT i j aux[i][j] -   (v1 - v2 - v4 + v3));   }   }   return;  }  // A function to answer the queries  static void answerQueries(Query q[] int m int BIT[][])  {   for (int i = 0; i < m; i++)   {   int x1 = q[i].x1 + 1;   int y1 = q[i].y1 + 1;   int x2 = q[i].x2 + 1;   int y2 = q[i].y2 + 1;   int ans = getSum(BIT x2 y2) -  getSum(BIT x2 y1 - 1) -   getSum(BIT x1 - 1 y2) +  getSum(BIT x1 - 1 y1 - 1);   System.out.printf('Query(%d %d %d %d) = %dn'   q[i].x1 q[i].y1 q[i].x2 q[i].y2 ans);   }   return;  }  // Driver Code public static void main(String[] args)  {   int mat[][] = { {1 2 3 4}   {5 3 8 1}   {4 6 7 5}   {2 4 8 9} };   // Create a 2D Binary Indexed Tree   int [][]BIT = new int[N + 1][N + 1];   construct2DBIT(mat BIT);   /* Queries of the form - x1 y1 x2 y2   For example the query- {1 1 3 2} means the sub-matrix-   y   /   3 | 1 2 3 4 Sub-matrix   2 | 5 3 8 1 {1132} --. 3 8 1   1 | 4 6 7 5 6 7 5   0 | 2 4 8 9   |   --|------ 0 1 2 3 ---. x   |     Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30   */  Query q[] = {new Query(1 1 3 2)   new Query(2 3 3 3)   new Query(1 1 1 1)};   int m = q.length;   answerQueries(q m BIT);  } }  // This code is contributed by 29AjayKumar 
Python3
'''Python3 program to implement 2D Binary Indexed Tree    2D BIT is basically a BIT where each element is another BIT.  Updating by adding v on (x y) means it's effect will be found  throughout the rectangle [(x y) (max_x max_y)]  and query for (x y) gives you the result of the rectangle  [(0 0) (x y)] assuming the total rectangle is  [(0 0) (max_x max_y)]. So when you query and update on  this BITyou have to be careful about how many times you are  subtracting a rectangle and adding it. Simple set union formula  works here.    So if you want to get the result of a specific rectangle  [(x1 y1) (x2 y2)] the following steps are necessary:    Query(x1y1x2y2) = getSum(x2 y2)-getSum(x2 y1-1) -   getSum(x1-1 y2)+getSum(x1-1 y1-1)    Here 'Query(x1y1x2y2)' means the sum of elements enclosed  in the rectangle with bottom-left corner's co-ordinates  (x1 y1) and top-right corner's co-ordinates - (x2 y2)    Constraints -> x1<=x2 and y1<=y2     /  y |   | --------(x2y2)   | | |   | | |   | | |   | ---------   | (x1y1)   |   |___________________________  (0 0) x-->    In this program we have assumed a square matrix. The  program can be easily extended to a rectangular one. ''' N = 4 # N-.max_x and max_y  # A structure to hold the queries  class Query: def __init__(self x1y1x2y2): self.x1 = x1; self.y1 = y1; self.x2 = x2; self.y2 = y2; # A function to update the 2D BIT  def updateBIT(BITxyval): while x <= N: # This loop update all the 1D BIT inside the  # array of 1D BIT = BIT[x]  while y <= N: BIT[x][y] += val; y += (y & -y) x += (x & -x) return; # A function to get sum from (0 0) to (x y)  def getSum(BITxy): sum = 0; while x > 0: # This loop sum through all the 1D BIT  # inside the array of 1D BIT = BIT[x]  while y > 0: sum += BIT[x][y]; y -= y&-y x -= x&-x return sum; # A function to create an auxiliary matrix  # from the given input matrix  def constructAux(mataux): # Initialise Auxiliary array to 0  for i in range(N + 1): for j in range(N + 1): aux[i][j] = 0 # Construct the Auxiliary Matrix  for j in range(1 N + 1): for i in range(1 N + 1): aux[i][j] = mat[N - j][i - 1]; return # A function to construct a 2D BIT  def construct2DBIT(matBIT): # Create an auxiliary matrix  aux = [None for i in range(N + 1)] for i in range(N + 1) : aux[i]= [None for i in range(N + 1)] constructAux(mat aux) # Initialise the BIT to 0  for i in range(1 N + 1): for j in range(1 N + 1): BIT[i][j] = 0; for j in range(1 N + 1): for i in range(1 N + 1): # Creating a 2D-BIT using update function  # everytime we/ encounter a value in the  # input 2D-array  v1 = getSum(BIT i j); v2 = getSum(BIT i j - 1); v3 = getSum(BIT i - 1 j - 1); v4 = getSum(BIT i - 1 j); # Assigning a value to a particular element  # of 2D BIT  updateBIT(BIT i j aux[i][j] - (v1 - v2 - v4 + v3)); return; # A function to answer the queries  def answerQueries(qmBIT): for i in range(m): x1 = q[i].x1 + 1; y1 = q[i].y1 + 1; x2 = q[i].x2 + 1; y2 = q[i].y2 + 1; ans = getSum(BIT x2 y2) -  getSum(BIT x2 y1 - 1) -  getSum(BIT x1 - 1 y2) +  getSum(BIT x1 - 1 y1 - 1); print('Query (' q[i].x1 ' ' q[i].y1 ' ' q[i].x2 ' '  q[i].y2 ') = ' ans sep = '') return; # Driver Code mat= [[1 2 3 4] [5 3 8 1] [4 6 7 5] [2 4 8 9]]; # Create a 2D Binary Indexed Tree  BIT = [None for i in range(N + 1)] for i in range(N + 1): BIT[i]= [None for i in range(N + 1)] for j in range(N + 1): BIT[i][j]=0 construct2DBIT(mat BIT);   ''' Queries of the form - x1 y1 x2 y2   For example the query- {1 1 3 2} means the sub-matrix-   y   /   3 | 1 2 3 4 Sub-matrix   2 | 5 3 8 1 {1132} --. 3 8 1   1 | 4 6 7 5 6 7 5   0 | 2 4 8 9   |   --|------ 0 1 2 3 ---. x   |     Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30     ''' q = [Query(1 1 3 2) Query(2 3 3 3) Query(1 1 1 1)]; m = len(q) answerQueries(q m BIT); # This code is contributed by phasing17 
C#
/* C# program to implement 2D Binary Indexed Tree  2D BIT is basically a BIT where each element is another BIT.  Updating by.Adding v on (x y) means it's effect will be found  throughout the rectangle [(x y) (max_x max_y)]  and query for (x y) gives you the result of the rectangle  [(0 0) (x y)] assuming the total rectangle is  [(0 0) (max_x max_y)]. So when you query and update on  this BITyou have to be careful about how many times you are  subtracting a rectangle and.Adding it. Simple set union formula  works here.  So if you want to get the result of a specific rectangle  [(x1 y1) (x2 y2)] the following steps are necessary:  Query(x1y1x2y2) = getSum(x2 y2)-getSum(x2 y1-1) -   getSum(x1-1 y2)+getSum(x1-1 y1-1)  Here 'Query(x1y1x2y2)' means the sum of elements enclosed  in the rectangle with bottom-left corner's co-ordinates  (x1 y1) and top-right corner's co-ordinates - (x2 y2)  Constraints -> x1<=x2 and y1<=y2   /  y |   | --------(x2y2)   | | |   | | |   | | |   | ---------   | (x1y1)   |   |___________________________  (0 0) x-->  In this program we have assumed a square matrix. The  program can be easily extended to a rectangular one. */ using System; class GFG {  static readonly int N = 4; // N-.max_x and max_y  // A structure to hold the queries  public class Query  {   public int x1 y1; // x and y co-ordinates of bottom left   public int x2 y2; // x and y co-ordinates of top right   public Query(int x1 int y1 int x2 int y2)  {  this.x1 = x1;  this.y1 = y1;  this.x2 = x2;  this.y2 = y2;  }   };  // A function to update the 2D BIT  static void updateBIT(int []BIT int x  int y int val)  {   for (; x <= N; x += (x & -x))   {   // This loop update all the 1D BIT inside the   // array of 1D BIT = BIT[x]   for (; y <= N; y += (y & -y))   BIT[xy] += val;   }   return;  }  // A function to get sum from (0 0) to (x y)  static int getSum(int []BIT int x int y)  {   int sum = 0;   for(; x > 0; x -= x&-x)   {   // This loop sum through all the 1D BIT   // inside the array of 1D BIT = BIT[x]   for(; y > 0; y -= y&-y)   {   sum += BIT[x y];   }   }   return sum;  }  // A function to create an auxiliary matrix  // from the given input matrix  static void constructAux(int []mat int []aux)  {   // Initialise Auxiliary array to 0   for (int i = 0; i <= N; i++)   for (int j = 0; j <= N; j++)   aux[i j] = 0;   // Construct the Auxiliary Matrix   for (int j = 1; j <= N; j++)   for (int i = 1; i <= N; i++)   aux[i j] = mat[N - j i - 1];   return;  }  // A function to construct a 2D BIT  static void construct2DBIT(int []mat  int []BIT)  {   // Create an auxiliary matrix   int []aux = new int[N + 1 N + 1];   constructAux(mat aux);   // Initialise the BIT to 0   for (int i = 1; i <= N; i++)   for (int j = 1; j <= N; j++)   BIT[i j] = 0;   for (int j = 1; j <= N; j++)   {   for (int i = 1; i <= N; i++)   {   // Creating a 2D-BIT using update function   // everytime we/ encounter a value in the   // input 2D-array   int v1 = getSum(BIT i j);   int v2 = getSum(BIT i j - 1);   int v3 = getSum(BIT i - 1 j - 1);   int v4 = getSum(BIT i - 1 j);   // Assigning a value to a particular element   // of 2D BIT   updateBIT(BIT i j aux[ij] -   (v1 - v2 - v4 + v3));   }   }   return;  }  // A function to answer the queries  static void answerQueries(Query []q int m int []BIT)  {   for (int i = 0; i < m; i++)   {   int x1 = q[i].x1 + 1;   int y1 = q[i].y1 + 1;   int x2 = q[i].x2 + 1;   int y2 = q[i].y2 + 1;   int ans = getSum(BIT x2 y2) -  getSum(BIT x2 y1 - 1) -   getSum(BIT x1 - 1 y2) +  getSum(BIT x1 - 1 y1 - 1);   Console.Write('Query({0} {1} {2} {3}) = {4}n'   q[i].x1 q[i].y1 q[i].x2 q[i].y2 ans);   }   return;  }  // Driver Code public static void Main(String[] args)  {   int []mat = { {1 2 3 4}   {5 3 8 1}   {4 6 7 5}   {2 4 8 9} };   // Create a 2D Binary Indexed Tree   int []BIT = new int[N + 1N + 1];   construct2DBIT(mat BIT);   /* Queries of the form - x1 y1 x2 y2   For example the query- {1 1 3 2} means the sub-matrix-   y   /   3 | 1 2 3 4 Sub-matrix   2 | 5 3 8 1 {1132} --. 3 8 1   1 | 4 6 7 5 6 7 5   0 | 2 4 8 9   |   --|------ 0 1 2 3 ---. x   |     Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30   */  Query []q = {new Query(1 1 3 2)   new Query(2 3 3 3)   new Query(1 1 1 1)};   int m = q.Length;   answerQueries(q m BIT);  } } // This code is contributed by Rajput-Ji 
JavaScript
<script> /* Javascript program to implement 2D Binary Indexed Tree    2D BIT is basically a BIT where each element is another BIT.  Updating by adding v on (x y) means it's effect will be found  throughout the rectangle [(x y) (max_x max_y)]  and query for (x y) gives you the result of the rectangle  [(0 0) (x y)] assuming the total rectangle is  [(0 0) (max_x max_y)]. So when you query and update on  this BITyou have to be careful about how many times you are  subtracting a rectangle and adding it. Simple set union formula  works here.    So if you want to get the result of a specific rectangle  [(x1 y1) (x2 y2)] the following steps are necessary:    Query(x1y1x2y2) = getSum(x2 y2)-getSum(x2 y1-1) -   getSum(x1-1 y2)+getSum(x1-1 y1-1)    Here 'Query(x1y1x2y2)' means the sum of elements enclosed  in the rectangle with bottom-left corner's co-ordinates  (x1 y1) and top-right corner's co-ordinates - (x2 y2)    Constraints -> x1<=x2 and y1<=y2     /  y |   | --------(x2y2)   | | |   | | |   | | |   | ---------   | (x1y1)   |   |___________________________  (0 0) x-->    In this program we have assumed a square matrix. The  program can be easily extended to a rectangular one. */ let N = 4; // N-.max_x and max_y  // A structure to hold the queries  class Query  {  constructor(x1y1x2y2)  {  this.x1 = x1;  this.y1 = y1;  this.x2 = x2;  this.y2 = y2;  } } // A function to update the 2D BIT  function updateBIT(BITxyval) {  for (; x <= N; x += (x & -x))   {   // This loop update all the 1D BIT inside the   // array of 1D BIT = BIT[x]   for (; y <= N; y += (y & -y))   BIT[x][y] += val;   }   return;  } // A function to get sum from (0 0) to (x y)  function getSum(BITxy) {  let sum = 0;     for(; x > 0; x -= x&-x)   {   // This loop sum through all the 1D BIT   // inside the array of 1D BIT = BIT[x]   for(; y > 0; y -= y&-y)   {   sum += BIT[x][y];   }   }   return sum;  } // A function to create an auxiliary matrix  // from the given input matrix  function constructAux(mataux) {  // Initialise Auxiliary array to 0   for (let i = 0; i <= N; i++)   for (let j = 0; j <= N; j++)   aux[i][j] = 0;     // Construct the Auxiliary Matrix   for (let j = 1; j <= N; j++)   for (let i = 1; i <= N; i++)   aux[i][j] = mat[N - j][i - 1];     return;  } // A function to construct a 2D BIT  function construct2DBIT(matBIT) {  // Create an auxiliary matrix   let aux = new Array(N + 1);  for(let i=0;i<(N+1);i++)  {  aux[i]=new Array(N+1);  }  constructAux(mat aux);     // Initialise the BIT to 0   for (let i = 1; i <= N; i++)   for (let j = 1; j <= N; j++)   BIT[i][j] = 0;     for (let j = 1; j <= N; j++)   {   for (let i = 1; i <= N; i++)   {   // Creating a 2D-BIT using update function   // everytime we/ encounter a value in the   // input 2D-array   let v1 = getSum(BIT i j);   let v2 = getSum(BIT i j - 1);   let v3 = getSum(BIT i - 1 j - 1);   let v4 = getSum(BIT i - 1 j);     // Assigning a value to a particular element   // of 2D BIT   updateBIT(BIT i j aux[i][j] -   (v1 - v2 - v4 + v3));   }   }   return;  } // A function to answer the queries  function answerQueries(qmBIT) {  for (let i = 0; i < m; i++)   {   let x1 = q[i].x1 + 1;   let y1 = q[i].y1 + 1;   let x2 = q[i].x2 + 1;   let y2 = q[i].y2 + 1;     let ans = getSum(BIT x2 y2) -  getSum(BIT x2 y1 - 1) -   getSum(BIT x1 - 1 y2) +  getSum(BIT x1 - 1 y1 - 1);     document.write('Query ('+q[i].x1+' ' +q[i].y1+' ' +q[i].x2+' ' +q[i].y2+') = ' +ans+'  
'
); } return; } // Driver Code let mat= [[1 2 3 4] [5 3 8 1] [4 6 7 5] [2 4 8 9]]; // Create a 2D Binary Indexed Tree let BIT = new Array(N + 1); for(let i=0;i<(N+1);i++) { BIT[i]=new Array(N+1); for(let j=0;j<(N+1);j++) { BIT[i][j]=0; } } construct2DBIT(mat BIT); /* Queries of the form - x1 y1 x2 y2 For example the query- {1 1 3 2} means the sub-matrix- y / 3 | 1 2 3 4 Sub-matrix 2 | 5 3 8 1 {1132} --. 3 8 1 1 | 4 6 7 5 6 7 5 0 | 2 4 8 9 | --|------ 0 1 2 3 ---. x | Hence sum of the sub-matrix = 3+8+1+6+7+5 = 30 */ let q = [new Query(1 1 3 2) new Query(2 3 3 3) new Query(1 1 1 1)]; let m = q.length; answerQueries(q m BIT); // This code is contributed by rag2127 </script>

Izvade
Query(1 1 3 2) = 30 Query(2 3 3 3) = 7 Query(1 1 1 1) = 6 

Laika sarežģītība:  

  • Gan UpdateBit (X Y val) funkcija, gan getum (x y) funkcija ņem o (log (n)*log (m)) laiku.
  • 2D bitu veidošana ņem O (nm log (n)*log (m)).
  • Tā kā katrā no vaicājumiem, kurus mēs saucam par Getum (X Y) funkciju, tāpēc atbildot uz visiem Q vaicājumiem O (q*log (n)*log (m)) laiks.
  • Tādējādi programmas kopējā laika sarežģītība ir O ((nm+q)*log (n)*log (m)) kur 
    N = visas matricas maksimālais x koordināts. 
    M = visas matricas maksimālais y koordināts. 
    Q = vaicājumu skaits.

Papildu telpa: O (nm), lai saglabātu bitu un palīgdarbību