Vzhľadom na spočiatku prázdnu sadu a niekoľko dotazov na ňom každé z nasledujúcich typov:
- Vložte „X“ sa vykonáva pomocou aktualizácie (1 0 10^6 x 1). Všimnite si, že koreň stromu sa odovzdáva index štartu, ktorý sa odovzdáva ako 0 a koncový index ako 10^6, takže všetky rozsahy, ktoré majú X, sú aktualizované.
- Odstrániť „X“ sa vykonáva pomocou aktualizácie (1 0 10^6 x -1). Všimnite si, že koreň stromu sa odovzdáva index štartu, ktorý sa odovzdáva ako 0 a koncový index ako 10^6, takže všetky rozsahy, ktoré majú X, sú aktualizované.
Príklad:
Input : Insert 1 Insert 4 Insert 7 Median Output : The first three queries should insert 1 4 and 7 into an empty set. The fourth query should return 4 (median of 1 4 7).
Zastaveného účelu predpokladáme nasledujúce, ale tieto predpoklady nie sú obmedzeniami tu diskutovanej metódy:
1. V každom prípade sú všetky prvky odlišné, čo sa nevyskytuje viac ako raz.
2. „Medián“ dotazu sa uskutočňuje iba vtedy, keď je v sade nepárny počet prvkov. (V prípade rovnomerných čísel budeme musieť urobiť dva dotazy na našom strome segmentu)
3. Prvky v rozsahu sada od 1 do +10^6.
Metóda 1 (naivná)
Pri naivnej implementácii môžeme urobiť prvé dva dotazy v O (1), ale posledný dotaz v O (max_elem), kde max_elem je maximálnym prvkom všetkých čias (vrátane odstránených prvkov).
Predpokladajme pole počítať [] (s veľkosťou 10^6 + 1) na udržanie počtu každého prvku v podskupine. Nasledujúce sú jednoduché a samoliečené algoritmy pre 3 dotazy:
Vložte dotaz X:
count[x]++; if (x > max_elem) max_elem = x; n++;
Odstrániť X dotaz:
if (count[x] > 0) count[x]--; n--;
Stredný dotaz:
sum = 0; i = 0; while( sum <= n / 2 ) { i++; sum += count[i]; } median = i; return median; Ilustrácia počtu polí [] predstavujúce množinu {1 4 7 8 9} Medián prvku je „7“:
„Medián“ dotazu má v úmysle nájsť (n + 1)/2 „1“ v poli v tomto prípade 3. „1“; Teraz robíme to isté pomocou segmentového stromu.
Metóda 2 (pomocou Strom )
Robíme a strom Na ukladanie súčtu intervalov, kde interval [A B] predstavuje počet prvkov prítomných v sade, ktorá je v súčasnosti v rozsahu [A B]. Napríklad, ak zvážime vyššie uvedený príklad dotazu (3 7), vráti 2 dotaz (4 4) vráti 1 dotaz (5 5) návraty 0.
Vložte a odstráňte dotazy sú jednoduché a obe sa dajú implementovať pomocou aktualizácie funkcie (int x int diff) (pridáva „dif“ v indexe „x“)
Algoritmus
// adds ‘diff’ at index ‘x’ update(node a b x diff) // If leaf node If a == b and a == x segmentTree[node] += diff // If non-leaf node and x lies in its range If x is in [a b] // Update children recursively update(2*node a (a + b)/2 x diff) update(2*node + 1 (a + b)/2 + 1 b x diff) // Update node segmentTree[node] = segmentTree[2 * node] + segmentTree[2 * node + 1]
Vyššie uvedená rekurzívna funkcia spustí O (log (max_elem)) (V tomto prípade je max_elem 10^6) a používa sa na vloženie a vymazanie nasledujúcimi hovormi:
Teraz funkcia na nájdenie indexu s KTH „1“, kde „k“ v tomto prípade bude vždy (n + 1) / 2 Toto bude fungovať podobne ako binárne vyhľadávanie, môžete si ho myslieť ako na rekurzívnu funkciu binárneho vyhľadávania na strome segmentu.
Urobme príklad, aby sme pochopili, že náš set má v súčasnosti prvky {1 4 7 8 9}, a preto je reprezentovaný nasledujúcim stromom segmentu.
Ak sme v uzle bez listov, sme si istí, že má obidve deti, uvidíme, či má ľavé dieťa viac alebo rovnaké číslo ako „k“, ak áno, sme si istí, že náš index leží v ľavom podstromi, ak má ľavica podstroma menší počet 1 ako k, potom sme si istí, že náš index leží v pravom podsube. Robíme to rekurzívne, aby sme dosiahli náš index a odtiaľ ho vraciame.
Algoritmus
1.findKth(node a b k) 2. If a != b 3. If segmentTree[ 2 * node ] >= k 4. return findKth(2*node a (a + b)/2 k) 5. else 6. return findKth(2*node + 1 (a + b)/2 + 1 b k - segmentTree[ 2 * node ]) 7. else 8. return a
Vyššie uvedená rekurzívna funkcia spustí O (log (max_elem)) .
// A C++ program to implement insert delete and // median queries using segment tree #include
// A Java program to implement insert delete and // median queries using segment tree import java.io.*; class GFG{ public static int maxn = 3000005; public static int max_elem = 1000000; // A global array to store segment tree. // Note: Since it is global all elements are 0. public static int[] segmentTree = new int[maxn]; // Update 'node' and its children in segment tree. // Here 'node' is index in segmentTree[] 'a' and // 'b' are starting and ending indexes of range stored // in current node. // 'diff' is the value to be added to value 'x'. public static void update(int node int a int b int x int diff) { // If current node is a leaf node if (a == b && a == x ) { // Add 'diff' and return segmentTree[node] += diff; return ; } // If current node is non-leaf and 'x' // is in its range if (x >= a && x <= b) { // Update both sub-trees left and right update(node * 2 a (a + b) / 2 x diff); update(node * 2 + 1 (a + b) / 2 + 1 b x diff); // Finally update current node segmentTree[node] = segmentTree[node * 2] + segmentTree[node * 2 + 1]; } } // Returns k'th node in segment tree public static int findKth(int node int a int b int k) { // Non-leaf node will definitely have both // children; left and right if (a != b) { // If kth element lies in the left subtree if (segmentTree[node * 2] >= k) { return findKth(node * 2 a (a + b) / 2 k); } // If kth one lies in the right subtree return findKth(node * 2 + 1 (a + b) / 2 + 1 b k - segmentTree[node * 2]); } // If at a leaf node return the index it stores // information about return (segmentTree[node] != 0) ? a : -1; } // Insert x in the set public static void insert(int x) { update(1 0 max_elem x 1); } // Delete x from the set public static void delete(int x) { update(1 0 max_elem x -1); } // Returns median element of the set // with odd cardinality only public static int median() { int k = (segmentTree[1] + 1) / 2; return findKth(1 0 max_elem k); } // Driver code public static void main(String[] args) { insert(1); insert(4); insert(7); System.out.println('Median for the set {147} = ' + median()); insert(8); insert(9); System.out.println('Median for the set {14789} = ' + median()); delete(1); delete(8); System.out.println('Median for the set {479} = ' + median()); } } // This code is contributed by avanitrachhadiya2155
# A Python3 program to implement insert delete and # median queries using segment tree maxn = 3000005 max_elem = 1000000 # A global array to store segment tree. # Note: Since it is global all elements are 0. segmentTree = [0 for i in range(maxn)] # Update 'node' and its children in segment tree. # Here 'node' is index in segmentTree[] 'a' and # 'b' are starting and ending indexes of range stored # in current node. # 'diff' is the value to be added to value 'x'. def update(node a b x diff): global segmentTree # If current node is a leaf node if (a == b and a == x ): # add 'diff' and return segmentTree[node] += diff return # If current node is non-leaf and 'x' is in its # range if (x >= a and x <= b): # update both sub-trees left and right update(node * 2 a (a + b)//2 x diff) update(node * 2 + 1 (a + b)//2 + 1 b x diff) # Finally update current node segmentTree[node] = segmentTree[node * 2] + segmentTree[node * 2 + 1] # Returns k'th node in segment tree def findKth(node a b k): global segmentTree # non-leaf node will definitely have both # children left and right if (a != b): # If kth element lies in the left subtree if (segmentTree[node * 2] >= k): return findKth(node * 2 a (a + b)//2 k) # If kth one lies in the right subtree return findKth(node * 2 + 1 (a + b)//2 + 1 b k - segmentTree[node * 2]) # if at a leaf node return the index it stores # information about return a if (segmentTree[node]) else -1 # insert x in the set def insert(x): update(1 0 max_elem x 1) # delete x from the set def delete(x): update(1 0 max_elem x -1) # returns median element of the set with odd # cardinality only def median(): k = (segmentTree[1] + 1) // 2 return findKth(1 0 max_elem k) # Driver code if __name__ == '__main__': insert(1) insert(4) insert(7) print('Median for the set {147} ='median()) insert(8) insert(9) print('Median for the set {14789} ='median()) delete(1) delete(8) print('Median for the set {479} ='median()) # This code is contributed by mohit kumar 29
// A C# program to implement insert delete // and median queries using segment tree using System; class GFG{ public static int maxn = 3000005; public static int max_elem = 1000000; // A global array to store segment tree. // Note: Since it is global all elements are 0. public static int[] segmentTree = new int[maxn]; // Update 'node' and its children in segment tree. // Here 'node' is index in segmentTree[] 'a' and // 'b' are starting and ending indexes of range stored // in current node. // 'diff' is the value to be added to value 'x'. public static void update(int node int a int b int x int diff) { // If current node is a leaf node if (a == b && a == x) { // Add 'diff' and return segmentTree[node] += diff; return ; } // If current node is non-leaf and 'x' // is in its range if (x >= a && x <= b) { // Update both sub-trees left and right update(node * 2 a (a + b) / 2 x diff); update(node * 2 + 1 (a + b) / 2 + 1 b x diff); // Finally update current node segmentTree[node] = segmentTree[node * 2] + segmentTree[node * 2 + 1]; } } // Returns k'th node in segment tree public static int findKth(int node int a int b int k) { // Non-leaf node will definitely have both // children; left and right if (a != b) { // If kth element lies in the left subtree if (segmentTree[node * 2] >= k) { return findKth(node * 2 a (a + b) / 2 k); } // If kth one lies in the right subtree return findKth(node * 2 + 1 (a + b) / 2 + 1 b k - segmentTree[node * 2]); } // If at a leaf node return the index it // stores information about if (segmentTree[node] != 0) { return a; } else { return -1; } } // Insert x in the set public static void insert(int x) { update(1 0 max_elem x 1); } // Delete x from the set public static void delete(int x) { update(1 0 max_elem x -1); } // Returns median element of the set // with odd cardinality only public static int median() { int k = (segmentTree[1] + 1) / 2; return findKth(1 0 max_elem k); } // Driver code static public void Main() { insert(1); insert(4); insert(7); Console.WriteLine('Median for the set {147} = ' + median()); insert(8); insert(9); Console.WriteLine('Median for the set {14789} = ' + median()); delete(1); delete(8); Console.WriteLine('Median for the set {479} = ' + median()); } } // This code is contributed by rag2127
<script> // A Javascript program to implement insert delete and // median queries using segment tree let maxn = 3000005; let max_elem = 1000000; // A global array to store segment tree. // Note: Since it is global all elements are 0. let segmentTree = new Array(maxn); for(let i=0;i<maxn;i++) { segmentTree[i]=0; } // Update 'node' and its children in segment tree. // Here 'node' is index in segmentTree[] 'a' and // 'b' are starting and ending indexes of range stored // in current node. // 'diff' is the value to be added to value 'x'. function update(nodeabxdiff) { // If current node is a leaf node if (a == b && a == x ) { // Add 'diff' and return segmentTree[node] += diff; return ; } // If current node is non-leaf and 'x' // is in its range if (x >= a && x <= b) { // Update both sub-trees left and right update(node * 2 a Math.floor((a + b) / 2) x diff); update(node * 2 + 1 Math.floor((a + b) / 2) + 1 b x diff); // Finally update current node segmentTree[node] = segmentTree[node * 2] + segmentTree[node * 2 + 1]; } } // Returns k'th node in segment tree function findKth(nodeabk) { // Non-leaf node will definitely have both // children; left and right if (a != b) { // If kth element lies in the left subtree if (segmentTree[node * 2] >= k) { return findKth(node * 2 a Math.floor((a + b) / 2) k); } // If kth one lies in the right subtree return findKth(node * 2 + 1 Math.floor((a + b) / 2) + 1 b k - segmentTree[node * 2]); } // If at a leaf node return the index it stores // information about return (segmentTree[node] != 0) ? a : -1; } // Insert x in the set function insert(x) { update(1 0 max_elem x 1); } // Delete x from the set function delet(x) { update(1 0 max_elem x -1); } // Returns median element of the set // with odd cardinality only function median() { let k = (segmentTree[1] + 1) / 2; return findKth(1 0 max_elem k); } // Driver code insert(1); insert(4); insert(7); document.write('Median for the set {147} = ' + median()+'
'); insert(8); insert(9); document.write('Median for the set {14789} = ' + median()+'
'); delet(1); delet(8); document.write('Median for the set {479} = ' + median()+'
'); // This code is contributed by unknown2108 </script>
Výstup:
Median for the set {147} = 4 Median for the set {14789} = 7 Median for the set {479} = 7
Záver:
Všetky tri dotazy spúšťajú O (log (max_elem)) V tomto prípade sa max_elem = 10^6, takže log (max_elem) sa rovná 20.
Strom segmentu používa O (max_elem) priestor.
Keby tam bol dotaz odstránenia, mohol by sa to urobiť aj slávnym algoritmom tu .