Proto Van Emde蟒蛇树|套装6 |查询：继任者和前任者

请首先参考Proto Van Emde Boas Tree上的所有先前文章。

后续查询过程：

1. 基本情况：对于大小为2的Proto-VEB，唯一的可能性是密钥为0，并且如果存在下一个密钥，则该密钥是其后继者或不存在后继者。因此，将应用相同的过程。
2. 递归：
   • 首先，我们将查看当前群集(表示存在查询关键字的群集)，如果存在大于查询关键字的任何关键字，那么我们将成为后继者，因此我们将其返回。
   • 如果不是上述情况，那么我们将递归调用摘要的后继过程，以在摘要中找到下一个真实值。如果摘要中没有下一个真值，则我们将返回-1表示没有更大的密钥存在。
   • 在上面的操作中，如果我们找到下一个真值，那么我们将找到该群集中存在的最小键，该键将成为查询键的后继。

请参见下图，以基本了解Successor查询的操作：

Predecessor的过程与后继者相同，但有一些小的更改，您应该尝试从上述描述中了解后继者，以了解后继者。请参阅下图以获得基本了解：

下面是实现：

// C++ implementation of the approach
#include 
using namespace std;
  
class Proto_Van_Emde_Boas {
public:
    // Total number of keys
    int universe_size;
  
    // Summary
    Proto_Van_Emde_Boas* summary;
  
    // Clusters array of Proto-VEB pointers
    vector clusters;
  
    int root(int u)
    {
        return int(sqrt(u));
    }
  
    // Function to return cluster numbers
    // in which key is present
    int high(int x)
    {
        return x / root(universe_size);
    }
  
    // Function to return position of x in cluster
    int low(int x)
    {
        return x % root(universe_size);
    }
  
    // Function to return the index from
    // cluster number and position
    int generate_index(int cluster, int position)
    {
        return cluster * root(universe_size) + position;
    }
  
    // Constructor
    Proto_Van_Emde_Boas(int size)
    {
        universe_size = size;
  
        // Base case
        if (size <= 2) {
  
            // Set summary to nullptr as there is no
            // more summary for size 2
            summary = nullptr;
  
            // Vector of two pointers
            // nullptr in starting
            clusters = vector(size, nullptr);
        }
        else {
  
            // Assiging Proto-VEB(sqrt(u)) to summary
            summary = new Proto_Van_Emde_Boas(root(size));
  
            // Creating array of Proto-VEB Tree pointers of size sqrt(u)
            // first all nullptrs are going to assign
            clusters = vector(root(size), nullptr);
  
            // Assigning Proto-VEB(sqrt(u)) to all its clusters
            for (int i = 0; i < root(size); i++) {
                clusters[i] = new Proto_Van_Emde_Boas(root(size));
            }
        }
    }
};
  
// Function that returns true if the
// key is present in the tree
bool isMember(Proto_Van_Emde_Boas* helper, int key)
{
  
    // If key is greater then universe_size then
    // returns false
    if (key >= helper->universe_size)
        return false;
  
    // If we reach at base case
    // the just return whether
    // pointer is nullptr then false
    // else return true
    if (helper->universe_size == 2) {
        return helper->clusters[key];
    }
    else {
  
        // Recursively go deep into the
        // level of Proto-VEB tree using its
        // cluster index and its position
        return isMember(helper->clusters[helper->high(key)],
                        helper->low(key));
    }
}
  
// Function to insert a key in the tree
void insert(Proto_Van_Emde_Boas*& helper, int key)
{
    // If we reach at base case
    // then assign Proto-VEB(1) in place
    // of nullptr
    if (helper->universe_size == 2) {
        helper->clusters[key] = new Proto_Van_Emde_Boas(1);
    }
    else {
  
        // Recursively using index of cluster and its
        // position in cluster
        insert(helper->clusters[helper->high(key)],
               helper->low(key));
  
        // Also do the same recusion in summary VEB
        insert(helper->summary, helper->high(key));
    }
}
  
// Function to return the minimum key from the tree
int minimum(Proto_Van_Emde_Boas* helper)
{
    // Base case chooses the least key
    // present in the cluster
    if (helper->universe_size == 2) {
        if (helper->clusters[0]) {
            return 0;
        }
        else if (helper->clusters[1]) {
            return 1;
        }
  
        // No keys present then return -1
        return -1;
    }
    else {
  
        // Recursively find in summary for
        // first 1 present in Proto-VEB
        int minimum_cluster = minimum(helper->summary);
        int offset;
  
        // If no key is present in
        // the cluster then return -1
        if (minimum_cluster == -1) {
            return -1;
        }
        else {
  
            // Recursively find the position of the key
            // in the minimum_cluster
            offset = minimum(helper->clusters[minimum_cluster]);
  
            // Returns overall index of minimum key
            return helper->generate_index(minimum_cluster, offset);
        }
    }
}
  
// Function to return the maximum key from the tree
int maximum(Proto_Van_Emde_Boas* helper)
{
  
    // Return the maximum key present in
    // the cluster
    if (helper->universe_size == 2) {
        if (helper->clusters[1]) {
            return 1;
        }
        else if (helper->clusters[0]) {
            return 0;
        }
  
        // Return -1 if no keys present in the
        // cluster
        return -1;
    }
    else {
  
        // Recursively find the last 1 present
        // in the summary
        int maximum_cluster = maximum(helper->summary);
        int offset;
  
        // If no key is present in
        // the cluster then return -1
        if (maximum_cluster == -1) {
            return -1;
        }
        else {
  
            // Recursively find the position of the key
            // in the maximum_cluster
            offset = maximum(helper->clusters[maximum_cluster]);
            return helper->generate_index(maximum_cluster, offset);
        }
    }
}
  
// Function to return the successor of key in the tree
int successor(Proto_Van_Emde_Boas* helper, int key)
{
    // Base case, returns key greater than
    // our query key in the cluster if present
    // else returns -1
    if (helper->universe_size == 2) {
        if (key == 0 && helper->clusters[1])
            return 1;
        else
            return -1;
    }
    else {
  
        // Check if any key is greater than query key in the cluster
        int offset = successor(helper->clusters[helper->high(key)],
                               helper->low(key));
  
        // If it is present then return its index
        if (offset != -1)
            return helper->generate_index(helper->high(key), offset);
        else {
  
            // If no successor is present within the cluster then
            // go to the summmary and find the next summary with
            // key present(1) named successor_cluster
            int successor_cluster = successor(helper->summary,
                                              helper->high(key));
  
            // If no next 1 in the summary then return -1
            if (successor_cluster == -1)
                return -1;
            else {
  
                // Find the minimum key in the successor_cluster
                offset = minimum(helper->clusters[successor_cluster]);
  
                // Generate its index and return
                return helper->generate_index(successor_cluster, offset);
            }
        }
    }
}
  
// Function to return the predecessor of key in the tree
int predecessor(Proto_Van_Emde_Boas* helper, int key)
{
  
    // Base case, find smaller key present in
    // the cluster
    // If present else return -1
    if (helper->universe_size == 2) {
        if (key == 1 && helper->clusters[0])
            return 0;
        else
            return -1;
    }
    else {
  
        // Check if any key is lower than query key in the cluster
        int offset = predecessor(helper->clusters[helper->high(key)],
                                 helper->low(key));
  
        // If it is present then return its index
        if (offset != -1)
            return helper->generate_index(helper->high(key), offset);
        else {
  
            // If no predecessor is present within the cluster then
            // go to the summmary and find the next summary with
            // key present(1) named predecessor_cluster
            int predecessor_cluster = predecessor(helper->summary,
                                                  helper->high(key));
  
            // If no next 1 in the summary then return -1
            if (predecessor_cluster == -1)
                return -1;
            else {
  
                // Find the maximum key in the predecessor_cluster
                offset = maximum(helper->clusters[predecessor_cluster]);
  
                // Generate its index and return
                return helper->generate_index(predecessor_cluster, offset);
            }
        }
    }
}
  
// Function to delete a key from the tree
void pveb_delete(Proto_Van_Emde_Boas*& helper, int key)
{
  
    // Base case: If the key is present
    // then make it nullptr
    if (helper->universe_size == 2) {
        if (helper->clusters[key]) {
            delete helper->clusters[key];
            helper->clusters[key] = nullptr;
        }
    }
    else {
  
        // Recursive delete to reach at the base case
        pveb_delete(helper->clusters[helper->high(key)], helper->low(key));
  
        bool isanyinCluster = false;
  
        // Iterate over the cluster of keys to check whether
        // any other key is present within that cluster
        // If yes then we should not update summary to 0
        // else update summary to 0
        for (int i = helper->high(key) * helper->root(helper->universe_size);
             i < (helper->high(key) + 1) * helper->root(helper->universe_size);
             i++) {
  
            // If member is present then break the loop
            if (isMember(helper->clusters[helper->high(key)], i)) {
                isanyinCluster = true;
                break;
            }
        }
  
        // If no member is present then
        // update summary to zero
        if (isanyinCluster == false) {
            pveb_delete(helper->summary, helper->high(key));
        }
    }
}
  
// Driver code
int main()
{
    Proto_Van_Emde_Boas* hello = new Proto_Van_Emde_Boas(16);
  
    cout << boolalpha;
  
    insert(hello, 2);
  
    insert(hello, 13);
  
    insert(hello, 3);
  
    cout << successor(hello, 3) << endl;
  
    cout << predecessor(hello, 13) << endl;
}

继任查询和前任查询的递归关系：

T(u)= T(u)= 2T( $\sqrt{u}$ ))+ O(log2( $\sqrt{u}$ ))

时间复杂度：O(log2(u)* log2(log2(u)))