Binārais indeksēts koks: diapazona atjaunināšana un diapazona vaicājumi
Dots masīvs arr[0..N-1]. Ir jāveic šādas darbības.
- atjauninājums (l r val) : pievienojiet “val” visiem masīva elementiem no [l r].
- iegūtRangeSum(l r) : atrodiet visu masīva elementu summu no [l r].
Sākotnēji visi masīva elementi ir 0. Vaicājumi var būt jebkurā secībā, t.i., pirms diapazona summas var būt daudz atjauninājumu.
Piemērs:
Ievade: N = 5 // {0 0 0 0 0}
Vaicājumi: atjauninājums: l = 0 r = 4 val = 2
atjauninājums: l = 3 r = 4 val = 3
getRangeSum : l = 2 r = 4Izvade: Diapazona [2 4] elementu summa ir 12
Paskaidrojums: Masīvs pēc pirmā atjaunināšanas kļūst par {2 2 2 2 2}
Masīvs pēc otrā atjaunināšanas kļūst par {2 2 2 5 5}
Naiva pieeja: Lai atrisinātu problēmu, izpildiet šādu ideju:
In iepriekšējā ziņa mēs apspriedām diapazona atjaunināšanu un punktu vaicājumu risinājumus, izmantojot BIT.
rangeUpdate(l r val) : Mēs pievienojam "val" elementam indeksā "l". Mēs atņemam "val" no elementa indeksā "r+1".
getElement(index) [vai getSum()]: mēs atgriežam elementu summu no 0 uz indeksu, ko var ātri iegūt, izmantojot BIT.
Mēs varam aprēķināt rangeSum(), izmantojot getSum() vaicājumus.
rangeSum(l r) = getSum(r) - getSum(l-1)Vienkāršs risinājums ir izmantot risinājumus, kas apspriesti iepriekšējā ziņa . Diapazona atjaunināšanas vaicājums ir tāds pats. Diapazona summas vaicājumu var sasniegt, veicot iegūšanas vaicājumu visiem diapazona elementiem.
Efektīva pieeja: Lai atrisinātu problēmu, izpildiet šādu ideju:
Mēs iegūstam diapazona summu, izmantojot prefiksu summas. Kā pārliecināties, ka atjaunināšana tiek veikta tā, lai prefiksu summu varētu veikt ātri? Apsveriet situāciju, kurā prefiksa summa [0 k] (kur 0 <= k < n) is needed after range update on the range [l r]. Three cases arise as k can possibly lie in 3 regions.
- 1. gadījums : 0 < k < l
- Atjaunināšanas vaicājums neietekmēs summas vaicājumu.
- 2. gadījums : l <= k <= r
- Apsveriet piemēru: pievienojiet 2 diapazonam [2 4], rezultāts būtu šāds masīvs: 0 0 2 2 2
Ja k = 3, summa no [0 k] = 4Kā iegūt šo rezultātu?
Vienkārši pievienojiet val no l th indekss uz k th rādītājs. Pēc atjaunināšanas vaicājuma summa tiek palielināta par "val*(k) - val*(l-1)".
- 3. gadījums : k > r
- Šajā gadījumā mums jāpievieno 'val' no l th indekss uz r th rādītājs. Summa tiek palielināta par 'val*r – val*(l-1)' atjaunināšanas vaicājuma dēļ.
Novērojumi:
1. gadījums: ir vienkārša, jo summa paliktu tāda pati kā pirms atjaunināšanas.
2. gadījums: Summa tika palielināta par val*k - val*(l-1). Mēs varam atrast “val”, tas ir līdzīgs i atrašanai th elements iekšā diapazona atjauninājumu un punktu vaicājuma rakstu . Tāpēc mēs uzturam vienu BIT diapazona atjaunināšanai un punktu vaicājumiem, šis BIT būs noderīgs, lai atrastu vērtību k th rādītājs. Tagad val * k tiek aprēķināts, kā rīkoties ar papildu terminu val*(l-1)?
Lai apstrādātu šo papildu termiņu, mēs uzturam vēl vienu BIT (BIT2). Atjaunināt val * (l-1) pie l th indekss, tāpēc, veicot getSum vaicājumu BIT2, rezultāts būs val*(l-1).
3. gadījums: 3. gadījuma summa tika palielināta ar 'val*r - val *(l-1)', šī vārda vērtību var iegūt, izmantojot BIT2. Tā vietā, lai pievienotu, mēs atņemam 'val*(l-1) - val*r', jo mēs varam iegūt šo vērtību no BIT2, pievienojot val*(l-1), kā mēs to darījām 2. gadījumā un atņemot val*r katrā atjaunināšanas darbībā.
Atjaunināt vaicājumu
Atjauninājums (BITree1 l val)
Atjauninājums (BITree1 r+1 -val)
UpdateBIT2(BITree2 l val*(l-1))
UpdateBIT2 (BITree2 r+1 -val*r)Diapazona summa
getSum(BITTree1 k) *k) — getSum(BITTree2 k)
Lai atrisinātu problēmu, veiciet tālāk norādītās darbības.
- Izveidojiet divus bināro indeksu kokus, izmantojot doto funkciju constructBITree()
- Lai atrastu summu noteiktā diapazonā, izsauciet funkciju rangeSum() ar parametriem kā norādīto diapazonu un bināri indeksētiem kokiem
- Izsaukt funkcijas summu, kas atgriezīs summu diapazonā [0 X]
- Atdeves summa(R) — summa(L-1)
- Šīs funkcijas ietvaros izsauciet funkciju getSum(), kas atgriezīs masīva summu no [0 X]
- Atgriezt getSum(Tree1 x) * x - getSum(tree2 x)
- Funkcijā getSum() izveidojiet veselu skaitļu summu, kas vienāda ar nulli, un palieliniet indeksu par 1
- Kamēr indekss ir lielāks par nulli, palieliniet summu par Tree[index]
- Samaziniet indeksu par (index & (-index)), lai pārvietotu indeksu uz koka vecākmezglu
- Atdeves summa
- Izdrukājiet summu dotajā diapazonā
Zemāk ir aprakstīta iepriekš minētās pieejas īstenošana:
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>
Izvade
Sum of elements from [14] is 50
Laika sarežģītība : O(q * log(N)), kur q ir vaicājumu skaits.
Palīgtelpa: O(N)
