Requisito previo: Fenwick Tree
Sabemos que para responder consultas de suma de rango en una array 1-D de manera eficiente, el árbol indexado binario (o Fenwick Tree) es la mejor opción (incluso mejor que el árbol de segmentos debido a que requiere menos memoria y es un poco más rápido que el árbol de segmentos). ).
¿Podemos responder consultas de suma de subarrays de manera eficiente utilizando el árbol indexado binario?
La respuesta es sí . Esto es posible utilizando un BIT 2D que no es más que una array de BIT 1D.
Algoritmo:
Consideramos el siguiente ejemplo. Supongamos que tenemos que encontrar la suma de todos los números dentro del área resaltada.
Asumimos el origen de la array en la parte inferior – O. Luego, un BIT 2D explota el hecho de que-
Sum under the marked area = Sum(OB) - Sum(OD) -
Sum(OA) + Sum(OC)
En nuestro programa, usamos la función getSum(x, y) que encuentra la suma de la array de (0, 0) a (x, y).
De ahí la siguiente fórmula:
Sum under the marked area = Sum(OB) - Sum(OD) -
Sum(OA) + Sum(OC)
The above formula gets reduced to,
Query(x1,y1,x2,y2) = getSum(x2, y2) -
getSum(x2, y1-1) -
getSum(x1-1, y2) +
getSum(x1-1, y1-1)
donde,
x1, y1 = coordenadas x e y de C
x2, y2 = coordenadas x e y de B
La función actualizarBIT(x, y, val) actualiza todos los elementos bajo la región – (x, y) a (N, M ) donde,
N = máxima coordenada X de toda la array.
M = máxima coordenada Y de toda la array.
El resto del procedimiento es bastante similar al del árbol indexado binario 1D. A continuación se muestra la implementación en C++ del árbol indexado 2D
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 BIT,you 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(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
getSum(x1-1, y2)+getSum(x1-1, y1-1)
Here 'Query(x1,y1,x2,y2)' 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 |
| --------(x2,y2)
| | |
| | |
| | |
| ---------
| (x1,y1)
|
|___________________________
(0, 0) x-->
In this program we have assumed a square matrix. The
program can be easily extended to a rectangular one. */
#include<bits/stdc++.h>
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) = %d\n",
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 {1,1,3,2} ---> 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 BIT,you 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(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
getSum(x1-1, y2)+getSum(x1-1, y1-1)
Here 'Query(x1,y1,x2,y2)' 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 |
| --------(x2,y2)
| | |
| | |
| | |
| ---------
| (x1,y1)
|
|___________________________
(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) = %d\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 + 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 {1,1,3,2} --. 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 BIT,you 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(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
getSum(x1-1, y2)+getSum(x1-1, y1-1)
Here 'Query(x1,y1,x2,y2)' 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 |
| --------(x2,y2)
| | |
| | |
| | |
| ---------
| (x1,y1)
|
|___________________________
(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, x1,y1,x2,y2):
self.x1 = x1;
self.y1 = y1;
self.x2 = x2;
self.y2 = y2;
# A function to update the 2D BIT
def updateBIT(BIT,x,y,val):
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(BIT,x,y):
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(mat,aux):
# 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(mat,BIT):
# 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(q,m,BIT):
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 {1,1,3,2} --. 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 BIT,you 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(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
getSum(x1-1, y2)+getSum(x1-1, y1-1)
Here 'Query(x1,y1,x2,y2)' 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 |
| --------(x2,y2)
| | |
| | |
| | |
| ---------
| (x1,y1)
|
|___________________________
(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[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);
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 + 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 {1,1,3,2} --. 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 BIT,you 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(x1,y1,x2,y2) = getSum(x2, y2)-getSum(x2, y1-1) -
getSum(x1-1, y2)+getSum(x1-1, y1-1)
Here 'Query(x1,y1,x2,y2)' 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 |
| --------(x2,y2)
| | |
| | |
| | |
| ---------
| (x1,y1)
|
|___________________________
(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(x1,y1,x2,y2)
{
this.x1 = x1;
this.y1 = y1;
this.x2 = x2;
this.y2 = y2;
}
}
// A function to update the 2D BIT
function updateBIT(BIT,x,y,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)
function getSum(BIT,x,y)
{
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(mat,aux)
{
// 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(mat,BIT)
{
// 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(q,m,BIT)
{
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+"<br>");
}
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 {1,1,3,2} --. 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>
Producción
Query(1, 1, 3, 2) = 30 Query(2, 3, 3, 3) = 7 Query(1, 1, 1, 1) = 6
Complejidad del tiempo:
- Tanto la función updateBIT(x, y, val) como la función getSum(x, y) tardan O(log(NM)).
- La construcción del BIT 2D requiere O(NM log(NM)).
- Dado que en cada una de las consultas llamamos a la función getSum(x, y), por lo que responder a todas las consultas Q lleva O(Q.log(NM)) tiempo.
Por lo tanto, la complejidad de tiempo general del programa es O((NM+Q).log(NM)) donde,
N = máxima coordenada X de toda la array.
M = máxima coordenada Y de toda la array.
Q = Número de consultas.
Espacio auxiliar: O(NM) para almacenar el BIT y la array auxiliar
Referencias: https://www.topcoder.com/community/data-science/data-science-tutorials/binary-indexed-trees/
Este artículo es una contribución de Rachit Belwariar . Si te gusta GeeksforGeeks y te gustaría contribuir, también puedes escribir un artículo usando write.geeksforgeeks.org o enviar tu artículo por correo a review-team@geeksforgeeks.org. Vea su artículo que aparece en la página principal de GeeksforGeeks y ayude a otros Geeks.
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Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA