用C++写一个splay平衡树
时间: 2023-03-09 19:53:03 浏览: 137
我不是很熟悉C语言,但是我可以给你一些关于splay平衡树的基本知识:splay平衡树是一种二叉搜索树,它的特点是能够将最近使用的元素放在树的顶层,以提高查询效率。它的搜索过程和普通二叉搜索树类似,但是它会在搜索过程中进行调整,将最近访问的元素移动到树的根节点。
相关问题
帮我生成一段splay平衡树代码
### 回答1:
我可以为你生成一段splay平衡树代码: // 插入代码
void SplayTree::insert(int key)
{
//如果根为空,则新节点作为根节点
if (root == NULL)
{
root = new SplayNode(key);
return;
}
// 寻找节点
SplayNode *current = root, *parent = NULL;
while (current != NULL)
{
parent = current;
if (key < current->key)
current = current->left;
else if (key > current->key)
current = current->right;
else
break;
}
// 如果key不存在,则插入
if (key != parent->key)
{
if (key < parent->key)
parent->left = new SplayNode(key);
else
parent->right = new SplayNode(key);
// 将新插入节点作为根节点
splay(parent);
}
}
// 删除代码
void SplayTree::remove(int key)
{
SplayNode *node = search(key);
if (node == NULL)
return;
// 如果没有孩子节点,则直接删除
if (node->left == NULL && node->right == NULL)
{
if (node->parent == NULL)
{
root = NULL;
}
else
{
if (node->parent->left == node)
node->parent->left = NULL;
else
node->parent->right = NULL;
}
}
// 如果有一个孩子节点,则将其孩子节点替换
else if (node->left == NULL || node->right == NULL)
{
SplayNode *child = (node->left)? node->left: node->right;
if (node->parent == NULL)
root = child;
else
{
if (node->parent->left == node)
node->parent->left = child;
else
node->parent->right = child;
}
}
// 如果有两个孩子节点,则找到其后继节点,并将后继节点替换
else
{
SplayNode *successor = minimum(node->right);
if (node->parent == NULL)
root = successor;
else
{
if (node->parent->left == node)
node->parent->left = successor;
else
node->parent->right = successor;
}
successor->left = node->left;
}
delete(node);
}
### 回答2:
Splay平衡树是一种自适应的二叉搜索树,它可以通过对树中的节点进行旋转操作来保持树的平衡性。下面是一个简单的Splay平衡树的实现示例:
```
#include <iostream>
using namespace std;
// 树的节点结构
struct Node {
int key; // 节点的键值
Node* left; // 左子节点指针
Node* right; // 右子节点指针
};
// 旋转操作
Node* rotate(Node* node, bool isLeftRotate) {
Node* child = isLeftRotate ? node->right : node->left;
if (isLeftRotate) {
node->right = child->left;
child->left = node;
} else {
node->left = child->right;
child->right = node;
}
return child;
}
// 结点splay操作
Node* splay(Node* root, int key) {
if (root == nullptr || root->key == key)
return root;
if (key < root->key) { // 在左子树中splay
if (root->left == nullptr)
return root;
if (key < root->left->key) { // 左-左情况
root->left->left = splay(root->left->left, key);
root = rotate(root, true);
} else if (key > root->left->key) { // 左-右情况
root->left->right = splay(root->left->right, key);
if (root->left->right != nullptr)
root->left = rotate(root->left, false);
}
return (root->left == nullptr) ? root : rotate(root, true);
} else { // 在右子树中splay
if (root->right == nullptr)
return root;
if (key > root->right->key) { // 右-右情况
root->right->right = splay(root->right->right, key);
root = rotate(root, false);
} else if (key < root->right->key) { // 右-左情况
root->right->left = splay(root->right->left, key);
if (root->right->left != nullptr)
root->right = rotate(root->right, true);
}
return (root->right == nullptr) ? root : rotate(root, false);
}
}
// 插入节点
Node* insert(Node* root, int key) {
if (root == nullptr) {
Node* node = new Node();
node->key = key;
node->left = node->right = nullptr;
return node;
}
root = splay(root, key); // 将节点splay到根
if (key == root->key) // 节点已存在,则不插入
return root;
Node* node = new Node();
node->key = key;
if (key < root->key) {
node->right = root;
node->left = root->left;
root->left = nullptr;
} else {
node->left = root;
node->right = root->right;
root->right = nullptr;
}
return node;
}
int main() {
Node* root = nullptr;
root = insert(root, 50);
root = insert(root, 30);
root = insert(root, 70);
root = insert(root, 20);
root = insert(root, 40);
root = insert(root, 60);
root = insert(root, 80);
return 0;
}
```
这段代码实现了Splay平衡树的基本功能,包括节点旋转操作(rotate),节点splay操作(splay)和插入节点操作(insert)。你可以根据自己的需要在此基础上进一步完善和扩展平衡树的功能。
### 回答3:
可以为您提供一段用C++语言实现的Splay平衡树代码:
```cpp
#include <iostream>
using namespace std;
struct Node {
int key;
Node* left;
Node* right;
};
Node* createNode(int key) {
Node* newNode = new Node();
if(newNode) {
newNode->key = key;
newNode->left = newNode->right = NULL;
}
return newNode;
}
Node* rightRotate(Node* x) {
Node* y = x->left;
x->left = y->right;
y->right = x;
return y;
}
Node* leftRotate(Node* x) {
Node* y = x->right;
x->right = y->left;
y->left = x;
return y;
}
Node* splay(Node* root, int key) {
if (root == NULL || root->key == key)
return root;
if (root->key > key) {
if (root->left == NULL) return root;
if (root->left->key > key) {
root->left->left = splay(root->left->left, key);
root = rightRotate(root);
} else if (root->left->key < key) {
root->left->right = splay(root->left->right, key);
if (root->left->right != NULL)
root->left = leftRotate(root->left);
}
return (root->left == NULL) ? root : rightRotate(root);
} else {
if (root->right == NULL) return root;
if (root->right->key > key) {
root->right->left = splay(root->right->left, key);
if (root->right->left != NULL)
root->right = rightRotate(root->right);
} else if (root->right->key < key) {
root->right->right = splay(root->right->right, key);
root = leftRotate(root);
}
return (root->right == NULL) ? root : leftRotate(root);
}
}
Node* insert(Node* root, int key) {
if (root == NULL)
return createNode(key);
root = splay(root, key);
if (root->key == key)
return root;
Node* newNode = createNode(key);
if (root->key > key) {
newNode->right = root;
newNode->left = root->left;
root->left = NULL;
} else {
newNode->left = root;
newNode->right = root->right;
root->right = NULL;
}
return newNode;
}
void inOrder(Node* root) {
if (root != NULL) {
inOrder(root->left);
cout << root->key << " ";
inOrder(root->right);
}
}
int main() {
Node* root = NULL;
root = insert(root, 10);
root = insert(root, 20);
root = insert(root, 30);
root = insert(root, 40);
root = insert(root, 50);
cout << "中序遍历结果为: ";
inOrder(root);
return 0;
}
```
此代码实现了Splay平衡树的插入操作和中序遍历,您可以根据需要进行进一步扩展和修改。
帮我写一个c++的LCT
好的,下面是一个简单的 C++ LCT 实现,包含了 LCT 基本操作:
```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
// 节点结构体
struct Node {
int ch[2], fa, rev;
int val, sum;
} tr[N];
// 判断节点x是否是其父节点的左儿子
bool is_left(int x) {
return tr[tr[x].fa].ch[0] == x;
}
// 反转标记
void reverse(int x) {
swap(tr[x].ch[0], tr[x].ch[1]);
tr[x].rev ^= 1;
}
// 维护反转标记
void pushdown(int x) {
if (tr[x].rev) {
reverse(tr[x].ch[0]);
reverse(tr[x].ch[1]);
tr[x].rev = 0;
}
}
// 维护sum值
void pushup(int x) {
tr[x].sum = tr[tr[x].ch[0]].sum + tr[tr[x].ch[1]].sum + tr[x].val;
}
// 旋转操作
void rotate(int x) {
int y = tr[x].fa, z = tr[y].fa;
int k = is_left(x), w = tr[x].ch[k ^ 1];
tr[y].ch[k] = w; tr[w].fa = y;
tr[z].ch[is_left(y)] = x; tr[x].fa = z;
tr[x].ch[k ^ 1] = y; tr[y].fa = x;
pushup(y); pushup(x);
}
// 递归下传反转标记
void splay(int x) {
static int stk[N];
int top = 0, y = x;
stk[++top] = y;
while (!stk[top]) {
y = tr[y].fa;
stk[++top] = y;
}
while (top) pushdown(stk[top--]);
while (is_left(x)) {
int y = tr[x].fa, z = tr[y].fa;
if (is_left(y)) rotate(y);
rotate(x);
}
while (!is_left(x)) {
int y = tr[x].fa, z = tr[y].fa;
if (!is_left(y)) rotate(y);
rotate(x);
}
}
// 将x节点到根节点的路径变为一条链
void access(int x) {
for (int y = 0; x; y = x, x = tr[x].fa) {
splay(x);
tr[x].ch[1] = y;
pushup(x);
}
}
// 将x节点所在的树变为splay树的根节点
void make_root(int x) {
access(x);
splay(x);
reverse(x);
}
// 查询x节点所在的树的根节点
int find_root(int x) {
access(x);
splay(x);
while (tr[x].ch[0]) {
pushdown(x);
x = tr[x].ch[0];
}
splay(x);
return x;
}
// 连接x和y两个节点
void link(int x, int y) {
make_root(x);
tr[x].fa = y;
}
// 断开x和y两个节点
void cut(int x, int y) {
make_root(x);
if (find_root(y) == x && tr[y].fa == x && !tr[y].ch[0]) {
tr[y].fa = tr[x].ch[1] = 0;
pushup(x);
}
}
// 修改x节点的值为val
void modify(int x, int val) {
splay(x);
tr[x].val = val;
pushup(x);
}
// 查询x节点到y节点路径上的所有节点的权值和
int query(int x, int y) {
make_root(x);
access(y);
splay(y);
return tr[y].sum;
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; ++i) {
scanf("%d", &tr[i].val);
tr[i].sum = tr[i].val;
}
while (m--) {
int opt, x, y;
scanf("%d%d%d", &opt, &x, &y);
if (opt == 0) printf("%d\n", query(x, y));
else if (opt == 1) link(x, y);
else if (opt == 2) cut(x, y);
else if (opt == 3) modify(x, y);
}
return 0;
}
```
这里只是一个简单的 LCT 实现,对于一些细节和优化并没有处理。如果需要了解更多关于 LCT 的内容,可以参考一些经典题目,如 P3834、P3836、P4219 等。
相关推荐
![cpp](https://img-home.csdnimg.cn/images/20210720083646.png)
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)
![rar](https://img-home.csdnimg.cn/images/20210720083606.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![](https://csdnimg.cn/download_wenku/file_type_ask_c1.png)
![pdf](https://img-home.csdnimg.cn/images/20210720083512.png)
![rar](https://img-home.csdnimg.cn/images/20210720083606.png)
![zip](https://img-home.csdnimg.cn/images/20210720083736.png)