Árbol indexado binario: actualización de rango y consultas de rango
Dada una matriz arr[0..N-1]. Es necesario realizar las siguientes operaciones.
- actualizar(l r val) : Agregue 'val' a todos los elementos de la matriz desde [l r].
- obtenerSumaRango(l r) : Encuentre la suma de todos los elementos en la matriz de [l r].
Inicialmente, todos los elementos de la matriz son 0. Las consultas pueden realizarse en cualquier orden, es decir, puede haber muchas actualizaciones antes de la suma del rango.
Ejemplo:
Aporte: norte = 5 // {0 0 0 0 0}
Consultas: actualización: l = 0 r = 4 val = 2
actualización: l = 3 r = 4 val = 3
obtenerSumaRango: l = 2 r = 4Producción: La suma de los elementos del rango [2 4] es 12
Explicación: La matriz después de la primera actualización se convierte en {2 2 2 2 2}
La matriz después de la segunda actualización se convierte en {2 2 2 5 5}
Enfoque ingenuo: Para resolver el problema siga la siguiente idea:
En el publicación anterior Discutimos la actualización de rangos y las soluciones de consulta puntual utilizando BIT.
rangeUpdate(l r val): Agregamos 'val' al elemento en el índice 'l'. Restamos 'val' del elemento en el índice 'r+1'.
getElement(index) [o getSum()]: Devolvemos la suma de elementos desde 0 al índice que se puede obtener rápidamente usando BIT.
Podemos calcular rangeSum() usando consultas getSum().
rangoSuma(l r) = obtenerSuma(r) - obtenerSuma(l-1)Una solución sencilla es utilizar las soluciones discutidas en el publicación anterior . La consulta de actualización de rango es la misma. La consulta de suma de rango se puede lograr realizando una consulta de obtención de todos los elementos del rango.
Enfoque eficiente: Para resolver el problema siga la siguiente idea:
Obtenemos la suma del rango usando sumas de prefijo. ¿Cómo asegurarse de que la actualización se realice de manera que la suma del prefijo se pueda realizar rápidamente? Considere una situación en la que la suma del prefijo [0 k] (donde 0 <= k < n) is needed after range update on the range [l r]. Three cases arise as k can possibly lie in 3 regions.
- Caso 1 : 0 < k < l
- La consulta de actualización no afectará la consulta de suma.
- Caso 2 : yo <= k <= r
- Considere un ejemplo: Sume 2 al rango [2 4] y la matriz resultante sería: 0 0 2 2 2
Si k = 3 La suma de [0 k] = 4¿Cómo obtener este resultado?
Simplemente agregue el valor de l th índice de k th índice. La suma se incrementa en 'val*(k) - val*(l-1)' después de la consulta de actualización.
- Caso 3 : k > r
- Para este caso necesitamos agregar 'val' de l th índice de r th índice. La suma se incrementa en 'val*r – val*(l-1)' debido a una consulta de actualización.
Observaciones:
Caso 1: es simple ya que la suma seguiría siendo la misma que antes de la actualización.
Caso 2: La suma se incrementó en val*k - val*(l-1). Podemos encontrar 'val', es similar a encontrar i th elemento en artículo de consulta de puntos y actualización de rango . Por lo tanto, mantenemos un BIT para la actualización de rango y consultas de puntos. Este BIT será útil para encontrar el valor en k. th índice. Ahora se calcula val * k ¿cómo manejar el término adicional val*(l-1)?
Para gestionar este plazo adicional mantenemos otro BIT (BIT2). Actualizar val * (l-1) en l th index, por lo que cuando se realiza la consulta getSum en BIT2, el resultado será val*(l-1).
Caso 3: La suma en el caso 3 se incrementó en 'val*r - val *(l-1)'; el valor de este término se puede obtener usando BIT2. En lugar de sumar, restamos 'val*(l-1) - val*r' ya que podemos obtener este valor de BIT2 sumando val*(l-1) como hicimos en el caso 2 y restando val*r en cada operación de actualización.
Consulta de actualización
Actualización (BITree1 l val)
Actualización (BITree1 r+1 -val)
ActualizaciónBIT2(BITree2 l val*(l-1))
ActualizaciónBIT2(BITree2 r+1 -val*r)Suma de rango
obtenerSuma(ÁrbolBIT1 k) *k) - obtenerSuma(ÁrbolBIT2 k)
Siga los pasos a continuación para resolver el problema:
- Cree los dos árboles de índice binario usando la función dada constructBITree()
- Para encontrar la suma en un rango determinado, llame a la función rangeSum() con parámetros como el rango dado y árboles indexados binarios.
- Llame a una función suma que devolverá una suma en el rango [0 X]
- Devolver suma(R) - suma(L-1)
- Dentro de esta función llame a la función getSum() que devolverá la suma de la matriz de [0 X]
- Devuelve obtenerSuma(Árbol1 x) * x - obtenerSuma(árbol2 x)
- Dentro de la función getSum() cree una suma entera igual a cero y aumente el índice en 1
- Mientras el índice sea mayor que cero aumente la suma en Árbol[índice]
- Disminuya el índice en (index & (-index)) para mover el índice al nodo principal en el árbol
- Suma de devolución
- Imprime la suma en el rango dado
A continuación se muestra la implementación del enfoque anterior:
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 ( BITree index ) { 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 ( BITree n index 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] function sum ( x BITTree1 BITTree2 ) { return ( getSum ( BITTree1 x ) * x ) - getSum ( BITTree2 x ); } function updateRange ( BITTree1 BITTree2 n val l 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 ); } function rangeSum ( l r BITTree1 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 ); } 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>
Producción
Sum of elements from [14] is 50
Complejidad del tiempo : O(q * log(N)) donde q es el número de consultas.
Espacio Auxiliar: EN)
