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)netīrs baļķisVienkāršs 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 lthindekss uz kthrādītājs. Pēc atjaunināšanas vaicājuma summa tiek palielināta par "val*(k) - val*(l-1)".
- 3. gadījums : k > r
- Šajā gadījumā mums jāpievieno 'val' no lthindekss uz rthrādītājs. 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 atrašanaithelements iekšā diapazona atjauninājumu un punktu vaicājuma rakstu . Tāpēc mēs uzturam vienu BIT diapazona atjaunināšanai un punktu vaicājumiem, šis BIT būs noderīgs, lai atrastu vērtību kthrādītājs. 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 lthindekss, 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
tīmekļa pārlūkprogrammas iestatījumigetSum(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
// 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 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# 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
<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>
Izvade
Sum of elements from [14] is 50
Laika sarežģītība : O(q * log(N)), kur q ir vaicājumu skaits.
Palīgtelpa: O(N)