Dvejetainis indeksuotas medis: diapazono atnaujinimas ir diapazono užklausos
Duotas masyvas arr[0..N-1]. Reikia atlikti šias operacijas.
- atnaujinimas (l r val) : pridėkite „val“ prie visų masyvo elementų iš [l r].
- gautiRangeSum(l r) : Raskite visų masyvo elementų sumą iš [l r].
Iš pradžių visi masyvo elementai yra 0. Užklausos gali būti bet kokia tvarka, ty gali būti daug atnaujinimų prieš diapazono sumą.
Pavyzdys:
Įvestis: N = 5 // {0 0 0 0 0}
Užklausos: atnaujinimas: l = 0 r = 4 val = 2
atnaujinimas: l = 3 r = 4 val = 3
gautiRangeSum : l = 2 r = 4Išvestis: Diapazono [2 4] elementų suma yra 12
Paaiškinimas: Masyvas po pirmojo atnaujinimo tampa {2 2 2 2 2}
Masyvas po antrojo atnaujinimo tampa {2 2 2 5 5}
Naivus požiūris: Norėdami išspręsti problemą, vadovaukitės toliau pateikta idėja:
Į ankstesnis įrašas aptarėme diapazono atnaujinimo ir taško užklausų sprendimus naudojant BIT.
rangeUpdate(l r val) : Prie indekso 'l' elemento pridedame 'val'. Mes atimame „val“ iš elemento, esančio indekse „r+1“.
getElement(index) [arba getSum()]: grąžiname elementų sumą nuo 0 iki indekso, kurią galima greitai gauti naudojant BIT.
Mes galime apskaičiuoti rangeSum() naudodami getSum() užklausas.
rangeSum(l r) = getSum(r) - getSum(l-1)Paprastas sprendimas yra naudoti sprendimus, aptartus ankstesnis įrašas . Diapazono atnaujinimo užklausa yra tokia pati. Diapazono sumos užklausą galima pasiekti atlikus visų diapazono elementų gavimo užklausą.
Efektyvus požiūris: Norėdami išspręsti problemą, vadovaukitės toliau pateikta idėja:
Mes gauname diapazono sumą naudodami priešdėlių sumas. Kaip įsitikinti, kad naujinimas atliktas taip, kad prefikso sumą būtų galima atlikti greitai? Apsvarstykite situaciją, kai priešdėlio suma [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 atvejis : 0 < k < l
- Atnaujinimo užklausa neturės įtakos sumos užklausai.
- 2 atvejis : l <= k <= r
- Apsvarstykite pavyzdį: Pridėkite 2 prie diapazono [2 4], gautas masyvas būtų toks: 0 0 2 2 2
Jei k = 3, suma iš [0 k] = 4Kaip gauti šį rezultatą?
Tiesiog pridėkite val nuo l th indeksas į k th indeksas. Po atnaujinimo užklausos suma padidinama „val*(k) – val*(l-1)“.
- 3 atvejis : k > r
- Šiuo atveju turime pridėti „val“ iš l th indeksas į r th indeksas. Suma padidinama 'val*r – val*(l-1)' dėl atnaujinimo užklausos.
Pastebėjimai:
1 atvejis: yra paprasta, nes suma išliks tokia pati, kokia buvo prieš atnaujinimą.
2 atvejis: Suma padidinta val*k – val*(l-1). Galime rasti „val“, tai panašu į i radimą th elementas viduje diapazono atnaujinimo ir taško užklausos straipsnis . Taigi mes palaikome vieną BIT diapazono atnaujinimui ir taškų užklausoms, šis BIT bus naudingas ieškant k reikšmę th indeksas. Dabar apskaičiuojamas val * k, kaip tvarkyti papildomą terminą val*(l-1)?
Siekdami apdoroti šį papildomą terminą, palaikome kitą BIT (BIT2). Atnaujinti val * (l-1) adresu l th indeksas, todėl kai užklausa getSum bus atlikta BIT2, rezultatas bus val*(l-1).
3 atvejis: 3 atveju suma buvo padidinta 'val*r - val *(l-1)', šio termino reikšmę galima gauti naudojant BIT2. Užuot pridėję, atimame „val*(l-1) - val*r“, nes šią reikšmę galime gauti iš BIT2 pridėdami val*(l-1), kaip padarėme 2 atveju, ir atimdami val*r kiekvienoje atnaujinimo operacijoje.
Atnaujinti užklausą
Atnaujinimas (BITree1 l val)
Atnaujinimas (BITree1 r+1 -val)
AtnaujintiBIT2(BITree2 l val*(l-1))
AtnaujintiBIT2 (BITree2 r+1 -val*r)Diapazono suma
getSum(BITTree1 k) *k) - getSum(BITTree2 k)
Norėdami išspręsti problemą, atlikite toliau nurodytus veiksmus.
- Sukurkite du dvejetainius indekso medžius naudodami nurodytą funkciją constructBITree()
- Norėdami rasti sumą tam tikrame diapazone, iškvieskite funkciją rangeSum() su parametrais kaip nurodyta diapazone ir dvejetainiais indeksuotais medžiais
- Iškvieskite funkcijos sumą, kuri grąžins sumą diapazone [0 X]
- Grąžinimo suma(R) – suma(L-1)
- Šios funkcijos viduje iškvieskite funkciją getSum(), kuri grąžins masyvo sumą iš [0 X]
- Grąžinti getSum(tree1 x) * x - getSum(tree2 x)
- Funkcijos getSum() viduje sukurkite sveikojo skaičiaus sumą, lygią nuliui, ir padidinkite indeksą 1
- Jei indeksas yra didesnis nei nulis, padidinkite sumą Tree[index]
- Sumažinkite indeksą (index & (-index)), kad perkeltumėte indeksą į pagrindinį medžio mazgą
- Grąžinimo suma
- Atspausdinkite sumą duotame diapazone
Toliau pateikiamas pirmiau minėto metodo įgyvendinimas:
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>
Išvestis
Sum of elements from [14] is 50
Laiko sudėtingumas : O(q * log(N)), kur q yra užklausų skaičius.
Pagalbinė erdvė: O(N)
