Binært indekseret træ: Range Update og Range Queries
Givet en matrix arr[0..N-1]. Følgende operationer skal udføres.
- opdatering(l r val) : Tilføj 'val' til alle elementerne i arrayet fra [l r].
- getRangeSum(l r) : Find summen af alle elementer i matrixen fra [l r].
Til at begynde med er alle elementerne i arrayet 0. Forespørgsler kan være i enhver rækkefølge, dvs. der kan være mange opdateringer før intervalsum.
Eksempel:
Input: N = 5 // {0 0 0 0 0}
Forespørgsler: opdatering: l = 0 r = 4 val = 2
opdatering: l = 3 r = 4 val = 3
getRangeSum : l = 2 r = 4Produktion: Summen af elementer i området [2 4] er 12
Forklaring: Array efter første opdatering bliver {2 2 2 2 2}
Array efter anden opdatering bliver {2 2 2 5 5}
Naiv tilgang: Følg nedenstående idé for at løse problemet:
I den tidligere indlæg vi diskuterede rækkeviddeopdatering og punktforespørgselsløsninger ved hjælp af BIT.
rangeUpdate(l r val): Vi tilføjer 'val' til elementet ved indeks 'l'. Vi trækker 'val' fra elementet ved indeks 'r+1'.
getElement(indeks) [eller getSum()]: Vi returnerer summen af elementer fra 0 til indeks, som hurtigt kan opnås ved hjælp af BIT.
Vi kan beregne rangeSum() ved hjælp af getSum()-forespørgsler.
rangeSum(l r) = getSum(r) - getSum(l-1)En simpel løsning er at bruge de løsninger, der er diskuteret i tidligere indlæg . Forespørgslen om opdatering af rækkevidde er den samme. Forespørgsel til intervalsum kan opnås ved at lave en get-forespørgsel for alle elementer i området.
Effektiv tilgang: Følg nedenstående idé for at løse problemet:
Vi får intervalsum ved hjælp af præfikssummer. Hvordan sikrer man sig, at opdateringen udføres på en måde, så præfikssum kan udføres hurtigt? Overvej en situation, hvor præfikssum [0 k] (hvor 0 <= k < n) is needed after range update on the range [l r]. Three cases arise as k can possibly lie in 3 regions.
- Tilfælde 1 : 0 < k < l
- Opdateringsforespørgslen vil ikke påvirke sumforespørgslen.
- Tilfælde 2 : l <= k <= r
- Overvej et eksempel: Tilføj 2 til området [2 4] den resulterende matrix ville være: 0 0 2 2 2
Hvis k = 3 Summen fra [0 k] = 4Hvordan får man dette resultat?
Du skal blot tilføje tallet fra l th indeks til k th indeks. Summen øges med 'val*(k) - val*(l-1)' efter opdateringsforespørgslen.
- Tilfælde 3 : k > r
- I dette tilfælde skal vi tilføje 'val' fra l th indeks til r th indeks. Summen øges med 'val*r – val*(l-1)' på grund af en opdateringsforespørgsel.
Observationer:
Case 1: er enkel, da summen ville forblive den samme, som den var før opdateringen.
Tilfælde 2: Summen blev forøget med val*k - val*(l-1). Vi kan finde 'val', det svarer til at finde i th element i områdeopdatering og punktforespørgselsartikel . Så vi vedligeholder én BIT til Range Update og Point Queries, denne BIT vil være nyttig til at finde værdien ved k th indeks. Nu beregnes val * k, hvordan man håndterer ekstra term val*(l-1)?
For at håndtere denne ekstra periode opretholder vi en anden BIT (BIT2). Opdater værdi * (l-1) kl th indeks, så når getSum-forespørgslen udføres på BIT2 vil resultatet give val*(l-1).
Case 3: Summen i tilfælde 3 blev forøget med 'val*r - val *(l-1)' værdien af denne term kan opnås ved hjælp af BIT2. I stedet for at addere trækker vi 'val*(l-1) - val*r', da vi kan få denne værdi fra BIT2 ved at tilføje val*(l-1), som vi gjorde i tilfælde 2 og trække val*r fra i hver opdateringsoperation.
Opdater forespørgsel
Opdatering (BITree1 l værdi)
Opdatering(BITree1 r+1 -værdi)
UpdateBIT2(BITree2 l værdi*(l-1))
UpdateBIT2(BITree2 r+1 -værdi*r)Rækkevidde Sum
getSum(BITTree1 k) *k) - getSum(BITTree2 k)
Følg nedenstående trin for at løse problemet:
- Opret de to binære indekstræer ved hjælp af den givne funktion constructBITree()
- For at finde summen i et givent område skal du kalde funktionen rangeSum() med parametre som det givne område og binære indekserede træer
- Kald en funktionssum, der returnerer en sum i området [0 X]
- Retursum(R) - sum(L-1)
- Inde i denne funktion kald funktionen getSum(), som returnerer summen af arrayet fra [0 X]
- Returner getSum(Træ1 x) * x - getSum(træ2 x)
- Inde i funktionen getSum() opret en heltalssum lig med nul og øg indekset med 1
- Mens indekset er større end nul, øges summen med træ[indeks]
- Formindsk indeks med (indeks & (-indeks)) for at flytte indekset til den overordnede node i træet
- Retursum
- Udskriv summen i det givne interval
Nedenfor er implementeringen af ovenstående tilgang:
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>
Produktion
Sum of elements from [14] is 50
Tidskompleksitet : O(q * log(N)) hvor q er antallet af forespørgsler.
Hjælpeplads: PÅ)
