セットに対する挿入、削除、中央値クエリを効率的に設計する
最初に空のセットがあり、それに対するいくつかのクエリが与えられた場合、それぞれ次のタイプの可能性があります。
- 「x」の挿入は、update(1 0 10^6 x 1) を使用して行われます。 x を持つすべての範囲が更新されるように、ツリーのルートには開始インデックスが 0 として渡され、終了インデックスが 10^6 として渡されることに注意してください。
- 「x」の削除は、update(1 0 10^6 x -1) を使用して行われます。 x を持つすべての範囲が更新されるように、ツリーのルートには開始インデックスが 0 として渡され、終了インデックスが 10^6 として渡されることに注意してください。
例：
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).
説明の目的で次のことを仮定しますが、これらの仮定はここで説明する方法の制限ではありません。
1. どのような場合でも、すべての要素は個別です。つまり、どの要素も複数回出現しません。
2. 「中央値」クエリは、セット内に奇数の要素がある場合にのみ作成されます。(偶数の場合、セグメント ツリーで 2 つのクエリを作成する必要があります)。
3. セット内の要素の範囲は 1 ～ +10^6 です。
方法 1 (ナイーブ)
単純な実装では、最初の 2 つのクエリは O(1) で実行できますが、最後のクエリは O(max_elem) で実行できます。ここで、max_elem はすべての時点の最大要素 (削除された要素を含む) です。
配列を想定してみましょう カウント[] (サイズ 10^6 + 1) を使用して、サブセット内の各要素の数を維持します。以下は、3 つのクエリの単純で説明不要のアルゴリズムです。
x クエリを挿入します。
count[x]++; if (x > max_elem) max_elem = x; n++;
x クエリを削除:
if (count[x] > 0) count[x]--; n--;
中央値クエリ:
sum = 0; i = 0; while( sum <= n / 2 ) { i++; sum += count[i]; } median = i; return median; 集合 {1 4 7 8 9} を表す配列 count[] の図、中央値の要素は「7」です。
「中央値」クエリは、配列内の (n + 1)/2 番目の「1」、この場合は 3 番目の「1」を見つけることを目的としています。次に、セグメント ツリーを使用して同じことを行います。
方法2(使用方法) セグメントツリー ）
私たちは、 セグメントツリー 間隔の合計を格納します。間隔 [a b] は、現在範囲 [a b] 内にあるセット内に存在する要素の数を表します。たとえば、上記の例を考えると、 query(3 7) は 2 を返します。 query(4 4) は 1 を返します。 query(5 5) は 0 を返します。
挿入クエリと削除クエリは単純で、両方とも関数 update(int x int diff) を使用して実装できます (インデックス 'x' に 'diff' を追加します)
アルゴリズム
// 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]
上記の再帰関数は以下で実行されます。 O( log( max_elem ) ) (この場合、max_elem は 10^6)、次の呼び出しで挿入と削除の両方に使用されます。
さて、k 番目の「1」を持つインデックスを見つける関数です。この場合の「k」は常に (n + 1) / 2 になります。これは二分探索とよく似た働きをします。セグメント ツリー上の再帰二分探索関数と考えることができます。
例を挙げて、セットには現在要素 { 1 4 7 8 9 } があり、したがって次のセグメント ツリーで表されることを理解しましょう。
非リーフノードにいる場合、両方の子があることを確信します。左の子が「k」以上の 1 を持っているかどうかを確認します。はいの場合、インデックスは左のサブツリーにあると確信します。そうでない場合、左のサブツリーの 1 の数が k よりも少ない場合、インデックスは右のサブツリーにあると確信します。これを再帰的に実行してインデックスに到達し、そこからインデックスを返します。
アルゴリズム
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
上記の再帰関数は以下で実行されます。 O( log(max_elem) ) 。
// A C++ program to implement insert delete and // median queries using segment tree #include #define maxn 3000005 #define max_elem 1000000 using namespace std ; // A global array to store segment tree. // Note: Since it is global all elements are 0. int segmentTree [ 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'. 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 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 ]) ? a : -1 ; } // insert x in the set void insert ( int x ) { update ( 1 0 max_elem x 1 ); } // delete x from the set void delete ( int x ) { update ( 1 0 max_elem x -1 ); } // returns median element of the set with odd // cardinality only int median () { int k = ( segmentTree [ 1 ] + 1 ) / 2 ; return findKth ( 1 0 max_elem k ); } // Driver code int main () { insert ( 1 ); insert ( 4 ); insert ( 7 ); cout < < 'Median for the set {147} = ' < < median () < < endl ; insert ( 8 ); insert ( 9 ); cout < < 'Median for the set {14789} = ' < < median () < < endl ; delete ( 1 ); delete ( 8 ); cout < < 'Median for the set {479} = ' < < median () < < endl ; return 0 ; }
Java // 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
Python3 # 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
C# // 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
JavaScript < 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 ( node a b x 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 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 ( node a b 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 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>
出力：
Median for the set {147} = 4 Median for the set {14789} = 7 Median for the set {479} = 7
結論：
3 つのクエリすべてが実行されます O( log(max_elem) ) この場合、max_elem = 10^6 なので、log(max_elem) はほぼ 20 に等しくなります。
セグメントツリーは以下を使用します O( max_elem ) 空間。
削除クエリが存在しなかった場合、この問題は有名なアルゴリズムでも解決できた可能性があります。 ここ 。
トップ記事
