Запросы составных чисел в подмассиве (с обновлениями точек)
Учитывая массив из N целых чисел, задача состоит в том, чтобы выполнить следующие две операции с данным массивом:
query(start, end) : Print the number of Composite numbers in the subarray from start to end
update(i, x) : update the value at index i to x, i.e arr[i] = x
Примеры :
Ввод : arr = {1, 12, 3, 5, 17, 9} Запрос 1: запрос (начало = 0, конец = 4) Запрос 2: обновление (i = 3, x = 6) Запрос 3: запрос (начало = 0, конец = 4) Выход : 1 2 Объяснение В запросе 1 подмассив [0 ... 4] имеет 1 составное число, а именно. {12} В запросе 2 значение индекса 3 обновляется до 6, массив arr теперь равен, {1, 12, 3, 6, 7, 9} В запросе 3 подмассив [0 ... 4] имеет 2 составных числа, а именно. {12, 6}
Поскольку нам необходимо обрабатывать как запросы диапазона, так и обновления точек, эффективным методом является использование дерева сегментов для решения проблемы. Для этой цели лучше всего подходит дерево сегментов.
Мы можем использовать Sieve of Eratosthenes для предварительной обработки всех простых чисел до максимального значения, которое arr i может принять, скажем MAX. Сложность по времени для этой операции будет O (MAX log (log (MAX))) .
Построение дерева сегментов:
Проблема может быть сведена к сумме подмассивов с использованием дерева сегментов.
Теперь мы можем построить дерево сегментов, в котором листовой узел представлен либо как 0 (если это простое число), либо как 1 (если это составное число).
Внутренние узлы дерева сегментов равны сумме его дочерних узлов, таким образом, узел представляет общие составные числа в диапазоне от L до R, где диапазон от L до R попадает под этот узел и поддерево под ним.
Обработка запросов и обновлений точек:
Всякий раз, когда мы получаем запрос от начала до конца, мы можем запросить в дереве сегментов сумму узлов в диапазоне от начала до конца , который, в свою очередь, представляет количество композитов в диапазоне от начала до конца.
Если нам нужно выполнить обновление точки и обновить значение с индекса i до x, мы проверяем следующие случаи:
Let the old value of arri be y and the new value be x.
Case 1: If x and y both are composites.
Count of composites in the subarray does not change, so we just update array and donot
modify the segment tree
Case 2: If x and y both are primes.
Count of composites in the subarray does not change, so we just update array and donot
modify the segment tree
Case 3: If y is composite but x is prime.
Count of composite numbers in the subarray decreases, so we update array and add -1 to every
range, the index i which is to be updated, is a part of in the segment tree
Case 4: If y is prime but x is composite.
Count of composite numbers in the subarray increases, so we update array and add 1 to every
range, the index i which is to be updated, is a part of in the segment tree
Below is the implementation of the above approach:
C++
// C++ program to find number of composite numbers in a // subarray and performing updates #include <bits/stdc++.h> using namespace std; #define MAX 1000 // Function to calculate primes upto MAX // using sieve of Eratosthenes void sieveOfEratosthenes( bool isPrime[]) { isPrime[1] = true ; for ( int p = 2; p * p <= MAX; p++) { // If prime[p] is not changed, then // it is a prime if (isPrime[p] == true ) { // Update all multiples of p for ( int i = p * 2; i <= MAX; i += p) isPrime[i] = false ; } } } // A utility function to get the middle // index from corner indexes. int getMid( int s, int e) { return s + (e - s) / 2; } /* A recursive function to get the number of composites in a given range of array indexes. The following are parameters for this function. st --> Pointer to segment tree index --> Index of current node in the segment tree. Initially 0 is passed as root is always at index 0. ss & se --> Starting and ending indexes of the segment represented by current node, i.e., st[index] qs & qe --> Starting and ending indexes of query range */ int queryCompositesUtil( int * st, int ss, int se, int qs, int qe, int index) { // If segment of this node is a part of given range, // then return the number of composites // in the segment if (qs <= ss && qe >= se) return st[index]; // If segment of this node is // outside the given range if (se < qs || ss > qe) return 0; // If a part of this segment // overlaps with the given range int mid = getMid(ss, se); return queryCompositesUtil(st, ss, mid, qs, qe, 2 * index + 1) + queryCompositesUtil(st, mid + 1, se, qs, qe, 2 * index + 2); } /* A recursive function to update the nodes which have the given index in their range. The following are parameters st, si, ss and se are same as getSumUtil() i --> index of the element to be updated. This index is in input array. diff --> Value to be added to all nodes which have i in range */ void updateValueUtil( int * st, int ss, int se, int i, int diff, int si) { // Base Case: If the input index // lies outside the range of // this segment if (i < ss || i > se) return ; // If the input index is in range of // this node, then update the value of // the node and its children st[si] = st[si] + diff; if (se != ss) { int mid = getMid(ss, se); updateValueUtil(st, ss, mid, i, diff, 2 * si + 1); updateValueUtil(st, mid + 1, se, i, diff, 2 * si + 2); } } // The function to update a value in input // array and segment tree. It uses updateValueUtil() // to update the value in segment tree void updateValue( int arr[], int * st, int n, int i, int new_val, bool isPrime[]) { // Check for erroneous input index if (i < 0 || i > n - 1) { printf ( "Invalid Input" ); return ; } int diff, oldValue; oldValue = arr[i]; // Update the value in array arr[i] = new_val; // Case 1: Old and new values both are primes if (isPrime[oldValue] && isPrime[new_val]) return ; // Case 2: Old and new values both composite if ((!isPrime[oldValue]) && (!isPrime[new_val])) return ; // Case 3: Old value was composite, new value is prime if (!isPrime[oldValue] && isPrime[new_val]) { diff = -1; } // Case 4: Old value was prime, new_val is composite if (isPrime[oldValue] && !isPrime[new_val]) { diff = 1; } // Update the values of nodes in segment tree updateValueUtil(st, 0, n - 1, i, diff, 0); } // Return number of composite numbers in range // from index qs (query start) to qe (query end). // It mainly uses queryCompositesUtil() void queryComposites( int * st, int n, int qs, int qe) { int compositesInRange = queryCompositesUtil(st, 0, n - 1, qs, qe, 0); cout << "Number of Composites in subarray from " << qs << " to " << qe << " = " << compositesInRange << "
" ; } // A recursive function that constructs Segment Tree // for array[ss..se]. // si is index of current node in segment tree st int constructSTUtil( int arr[], int ss, int se, int * st, int si, bool isPrime[]) { // If there is one element in array, check if it // is prime then store 1 in the segment tree else // store 0 and return if (ss == se) { // if arr[ss] is composite if (!isPrime[arr[ss]]) st[si] = 1; else st[si] = 0; return st[si]; } // If there are more than one elements, then recur // for left and right subtrees and store the sum // of the two values in this node int mid = getMid(ss, se); st[si] = constructSTUtil(arr, ss, mid, st, si * 2 + 1, isPrime) + constructSTUtil(arr, mid + 1, se, st, si * 2 + 2, isPrime); return st[si]; } /* Function to construct segment tree from given array. This function allocates memory for segment tree and calls constructSTUtil() to fill the allocated memory */ int * constructST( int arr[], int n, bool isPrime[]) { // Allocate memory for segment tree // Height of segment tree int x = ( int )( ceil (log2(n))); // Maximum size of segment tree int max_size = 2 * ( int ) pow (2, x) - 1; int * st = new int [max_size]; // Fill the allocated memory st constructSTUtil(arr, 0, n - 1, st, 0, isPrime); // Return the constructed segment tree return st; } // Driver Code int main() { int arr[] = { 1, 12, 3, 5, 17, 9 }; int n = sizeof (arr) / sizeof (arr[0]); /* Preprocess all primes till MAX. Create a boolean array "isPrime[0..MAX]". A value in prime[i] will finally be false if i is composite, else true. */ bool isPrime[MAX + 1]; memset (isPrime, true , sizeof isPrime); sieveOfEratosthenes(isPrime); // Build segment tree from given array int * st = constructST(arr, n, isPrime); // Query 1: Query(start = 0, end = 4) int start = 0; int end = 4; queryComposites(st, n, start, end); // Query 2: Update(i = 3, x = 6), i.e Update // a[i] to x int i = 3; int x = 6; updateValue(arr, st, n, i, x, isPrime); // Query 3: Query(start = 0, end = 4) start = 0; end = 4; queryComposites(st, n, start, end); return 0; } |
Python3
# Python3 program to find # number of composite numbers # in a subarray and performing # updates import math MAX = 1000 # Function to calculate primes # upto MAX using sieve of Eratosthenes def sieveOfEratosthenes(isPrime): isPrime[ 1 ] = True ; p = 2 while p * p < = MAX : # If prime[p] is not # changed, then # it is a prime if (isPrime[p] = = True ): # Update all multiples of p for i in range (p * 2 , MAX + 1 , p): isPrime[i] = False ; p + = 1 # A utility function to get # the middle index from # corner indexes. def getMid(s, e): return s + (e - s) / / 2 ; """ A recursive function to get the number of composites in a given range of array indexes. The following are parameters for this function. st --> Pointer to segment tree index --> Index of current node in the segment tree. Initially 0 is passed as root is always at index 0. ss & se --> Starting and ending indexes of the segment represented by current node, i.e., st[index] qs & qe --> Starting and ending indexes of query range """ def queryCompositesUtil(st, ss, se, qs, qe, index): # If segment of this node is a # part of given range, then # return the number of composites # in the segment if (qs < = ss and qe > = se): return st[index]; # If segment of this node is # outside the given range if (se < qs or ss > qe): return 0 ; # If a part of this segment # overlaps with the given range mid = getMid(ss, se); return (queryCompositesUtil(st, ss, mid, qs, qe, 2 * index + 1 ) + queryCompositesUtil(st, mid + 1 , se, qs, qe, 2 * index + 2 )); """ A recursive function to update the nodes which have the given index in their range. The following are parameters st, si, ss and se are same as getSumUtil() i --> index of the element to be updated. This index is in input array. diff --> Value to be added to all nodes which have i in range """ def updateValueUtil(st, ss, se, i, diff, si): # Base Case: If the input index # lies outside the range of # this segment if (i < ss or i > se): return ; # If the input index is in # range of this node, then # update the value of the # node and its children st[si] = st[si] + diff; if (se ! = ss): mid = getMid(ss, se); updateValueUtil(st, ss, mid, i, diff, 2 * si + 1 ); updateValueUtil(st, mid + 1 , se, i, diff, 2 * si + 2 ); # The function to update a value # in input array and segment tree. # It uses updateValueUtil() to # update the value in segment tree def updateValue(arr, st, n, i, new_val, isPrime): # Check for erroneous # input index if (i < 0 or i > n - 1 ): print ( "Invalid Input" ); return РЕКОМЕНДУЕМЫЕ СТАТЬИ |