本文介绍了一种基于红黑树的数据结构——最大重叠点（POM）树，用于求解多个区间最大重叠点的问题。文章详细阐述了POM树的设计思路，包括基础数据结构的选择、附加信息的维护、新操作的设计，并提供了完整的C++实现代码。
题目
同习题14.3-6我们依然运用数据结构扩张的四个步骤来解决这个问题。
步骤1：选取基础数据结构
我们选择红黑树作为基础数据结构来扩张，红黑树是这章的主题，扩张之后所得的树称之为最大重叠点（POM）树。由于要求得区间的最大重叠点，因而当然的，我们的节点必然包含一个域存储以该节点为根的子树的最大POM，即为pom域。
步骤2：附加信息
除了pom域之外，为了能够有效地维护pom域，我们还将增加另外三个域：对于节点x而言，p，当x为某区间的左端点时，p为1；若为右端点，则为-1；p_sum，记录以x为根的子树中所有p的和；max，记录以x为根的子树中p的最大值，其实max就是节点x中的pom值重叠的次数。
步骤3：对信息的维护
新增加的三个域都是为了能够有效维护pom而增加的，结合下面的更新某个节点信息的update函数我们来讨论如何维护这些信息。
void update(node *curr)
{
curr->p_sum = curr->left->p_sum + curr->p + curr->right->p_sum;
curr->max = Max(curr->left->max, curr->left->p_sum + curr->p,
curr->left->p_sum + curr->p + curr->right->max);
if (curr->left != nil && curr->max == curr->left->max) curr->pom = curr->left->pom;
else if (curr->right != nil &&
curr->max == curr->left->p_sum + curr->p + curr->right->max)curr->pom = curr->right->pom;
else curr->pom = curr->key;
}
函数第四行计算当前节点curr的max，最大值是左子树的max，左子树的p_sum与curr的p的和，以及左子树的p_sum、curr的p还有右子树的max的和，这三者之间的最大值。关于后两者的计算，可以这样理解：当树的大小从局部左子树扩张到当前节点，即包含节点时，左子树中可能既存在某些区间的左端点，也有右端点。那么对于当前节点而言，它势必不在这个区间中，因此不能够是左子树的max和curr的p值之和，而是p_sum与p的和，同理后后者也是如此。update函数最后的if语句则判断以x为根的子树中最大重叠点是哪个，这主要通过判断由上步算出的max与哪个值相等来确定。
明显地，该函数的执行时间为O(1)。同习题14.3-6，插入和删除过程的更新过程是不能做到自顶向下的，只能在确定位置之后，在自底向上去更新，查找时间是O(lgn)，向上更新也是，因而插入和删除的渐进时间并没有改变。
步骤4：设计新操作
除了update函数，还将增加findPom函数返回最大重叠点以及自底向上的更新函数keep。
OK，解决这四步，稍微对红黑树其他操作做些修改就可以构造出一颗最大重叠点树。下面是该树的C++实现代码：
/*****求多个区间的最大重叠点，姑且称为最大重叠点树，红黑树扩张
*增加了p,p_sum,max,pom等数据成员
*增加了update和keep函数
*微调insert，左右旋以及erase函数
*/
#include<iostream>
//#include<cstdlib>
using namespace std;
enum COLOR { red, black };//枚举，定义颜色
#define Max(i,j,k) (i > j ? (i > k ? i : k) : (j > k) ? j : k);
const int MAX = 0x7fffffff;
class node
{
private:
friend class POMTree;
node *parent;
node *left;
node *right;
int key;
int p;//左端点为1，右端点为-1
int p_sum;//以当前节点为根的子树中p的和
int max;//以当前节点为根的子树中p的和的最大值
int pom;//使其取到最大值的关键字,即此子树的最大重叠点
COLOR color;
node(){}//默认构造函数，只供创建nil时调用
public:
node(int k, COLOR c = red) :key(k), p_sum(0), max(0), pom(k), color(c),
parent(NULL), left(NULL), right(NULL){}
void print()const
{
printf("key:%-10d POM:%-10d max:%-10d p_sum:%-10d p:%-10d", key, pom, max, p_sum, p);
if (color == red) printf("red\n");
else printf("black\n");
}
//省略指针域的getter和setter
};
class POMTree
{
private:
static node *nil;//哨兵，静态成员，被整个POMTree类所共有
node *root;
POMTree(const POMTree&);//禁止复制构造
POMTree operator=(const POMTree&);//禁止赋值
void leftRotate(node*);//左旋
void rightRotate(node*);//右旋
void insertFixup(node*);//插入节点后红黑性质调整
void eraseFixup(node*);//删除节点后红黑性质调整
void update(node*);//更新节点信息
void keep(node*);//自底向上更新路径信息
public:
POMTree() :root(nil)
{
root->parent = nil;
root->left = nil;
root->right = nil;
root->color = black;
root->key = MAX; root->p = 0;
root->p_sum = 0; root->max = 0;
root->pom = MAX;
}
POMTree(node *rbt) :root(rbt){}//复制构造函数，用于创建子红黑树对象
void insert(int, bool);//插入
void create();//创建红黑树
void erase(int);//删除
node* locate(int)const;//查找
node* minMum()const;//最小值
node* maxMum()const;//最大值
node* successor(int)const;//找后继
node* predecessor(int)const;//前驱
void preTraversal()const;//先根遍历
void inTraversal()const;//中根遍历
void destroy();//销毁红黑树
void findPom()const { root->print(); }
bool empty()const{ return root == nil; }//判空
};
node *POMTree::nil = new node;//定义静态成员nil
void POMTree::update(node *curr)
{
curr->p_sum = curr->left->p_sum + curr->p + curr->right->p_sum;
curr->max = Max(curr->left->max, curr->left->p_sum + curr->p,
curr->left->p_sum + curr->p + curr->right->max);
if (curr->left != nil && curr->max == curr->left->max) curr->pom = curr->left->pom;
else if (curr->right != nil &&
curr->max == curr->left->p_sum + curr->p + curr->right->max)curr->pom = curr->right->pom;
else curr->pom = curr->key;
}
void POMTree::keep(node *curr)
{
while (curr != nil)
{
update(curr);
curr = curr->parent;
}
}
void POMTree::leftRotate(node *curr)
{
if (curr->right != nil)
{//存在右孩子时才能左旋
node *rchild = curr->right;
curr->right = rchild->left;
if (rchild->left != nil)
rchild->left->parent = curr;
rchild->parent = curr->parent;
if (curr->parent == nil)
root = rchild;
else if (curr == curr->parent->left)
curr->parent->left = rchild;
else curr->parent->right = rchild;
curr->parent = rchild;
rchild->left = curr;
rchild->p_sum = curr->p_sum;
rchild->max = curr->max;
rchild->pom = curr->pom;
update(curr);
}
}
void POMTree::rightRotate(node *curr)
{
if (curr->left != nil)
{//存在左孩子时才能右旋
node *lchild = curr->left;
curr->left = lchild->right;
if (lchild->right != nil)
lchild->right->parent = curr;
lchild->parent = curr->parent;
if (curr->parent == nil)
root = lchild;
else if (curr == curr->parent->left)
curr->parent->left = lchild;
else curr->parent->right = lchild;
lchild->right = curr;
curr->parent = lchild;
lchild->p_sum = curr->p_sum;
lchild->max = curr->max;
lchild->pom = curr->pom;
update(curr);
}
}
void POMTree::insert(int k, bool start)
{
node *pkey = new node(k),
*p = nil, *curr = root;
if (start)
{
pkey->p = 1;
pkey->p_sum = 1;
pkey->max = 1;
pkey->pom = k;
}
else
{
pkey->p = -1;
pkey->p_sum = -1;
pkey->max = 0;
pkey->pom = k;
}
while (curr != nil)
{//找插入位置
p = curr;//记住当前节点父亲
if (k < curr->key)//往左找
curr = curr->left;
else curr = curr->right;//向右找
}
pkey->parent = p;
if (p == nil)//插入的是第一个节点
root = pkey;
else if (k < p->key)
p->left = pkey;
else p->right = pkey;
pkey->left = pkey->right = nil;
keep(pkey->parent);
insertFixup(pkey);//调整
}
void POMTree::insertFixup(node *curr)
{
while (curr->parent->color == red)
{//父亲为红节点时才需要进入循环调整
if (curr->parent == curr->parent->parent->left)
{//父亲是祖父左孩子
node *uncle = curr->parent->parent->right;
if (uncle != nil && uncle->color == red)
{//情况1，叔叔节点存在且为红色
curr->parent->color = black;
uncle->color = black;
curr->parent->parent->color = red;
curr = curr->parent->parent;
}
else if (curr == curr->parent->right)
{//情况2，叔叔节点为黑色，且当前节点是父亲右孩子
curr = curr->parent;
leftRotate(curr);//将父节点左旋，以转变为情况3
}
else
{//情况3，叔叔节点为黑色，且当前节点是父亲左孩子
curr->parent->color = black;
curr->parent->parent->color = red;
rightRotate(curr->parent->parent);
}
}
else
{//父亲是祖父右孩子，与上面三种情况对称
node *uncle = curr->parent->parent->left;
if (uncle != nil && uncle->color == red)
{//情况1
curr->parent->color = black;
uncle->color = black;
curr->parent->parent->color = red;
curr = curr->parent->parent;
}
else if (curr == curr->parent->left)
{//情况2
curr = curr->parent;
rightRotate(curr);
}
else
{//情况3
curr->parent->color = black;
curr->parent->parent->color = red;
leftRotate(curr->parent->parent);
}
}
}
root->color = black;//跳出循环时将根节点置为黑色
}
void POMTree::create()
{
int low, high;
cout << "Enter element(s),CTRL+Z to end" << endl;//换行后CTRL+Z结束输入
while (cin >> low >> high)
{
insert(low, true);
insert(high, false);
}
cin.clear();
}
void POMTree::preTraversal()const
{
node *curr = root;
if (curr != nil)
{
curr->print();
POMTree LEFT(curr->left);//继续左子树先根遍历
LEFT.preTraversal();
POMTree RIGHT(curr->right);
RIGHT.preTraversal();
}
}
void POMTree::inTraversal()const
{
node *curr = root;
if (curr != nil)
{
POMTree LEFT(curr->left);
LEFT.inTraversal();
curr->print();
POMTree RIGHT(curr->right);//继续右子树中根遍历
RIGHT.inTraversal();
}
}
node* POMTree::successor(int k)const
{
node *curr = locate(k);
if (curr->right != nil)
{//若右子树不为空，则后继为右子树最小值
POMTree RIGHT(curr->right);
return RIGHT.minMum();
}
node *p = curr->parent;
while (p != nil && curr == p->right)
{//否则为沿右指针一直向上直到第一个拐弯处节点
curr = p;
p = p->parent;
}
return p;
}
node* POMTree::minMum()const
{
node *curr = root;
while (curr->left != nil)
curr = curr->left;
return curr;
}
node* POMTree::maxMum()const
{
node *curr = root;
while (curr->right != nil)
curr = curr->right;
return curr;
}
node* POMTree::predecessor(int k)const
{
node *curr = locate(k);
if (curr->left != nil)
{//若左子树不为空，则前驱为左子树最大值
POMTree LEFT(curr->left);
return LEFT.maxMum();
}
node *p = curr->parent;
while (p != nil && curr == p->left)
{//否则为沿左指针一直往上的第一个拐弯处节点
curr = p;
p = p->parent;
}
return p;
}
void POMTree::erase(int k)
{
node *curr = locate(k), *pdel, *child;
if (curr == nil)
{
cout << "Error:no data" << endl;
return;
}
if (curr->left == nil || curr->right == nil)//决定删除节点
pdel = curr;//若当前节点至多有一个孩子，则删除它
else pdel = successor(k);//否则若有两孩子，则删除其后继
node *pdel_parent = pdel->parent;//记下被删节点父亲
if (pdel->left != nil)//记下不为空的孩子
child = pdel->left;
else child = pdel->right;
child->parent = pdel_parent;
if (pdel_parent == nil)//若删除的是根节点
root = child;
else if (pdel == pdel_parent->left)//否则若被删节点是其父亲左孩子
pdel_parent->left = child;
else pdel_parent->right = child;
if (curr != pdel)
curr->key = pdel->key;//若被删的是后继，则将后继值赋给当前节点
keep(pdel_parent);
if (pdel->color == black)//被删节点为黑色时才调整
eraseFixup(child);
delete pdel;//释放所占内存
}
void POMTree::eraseFixup(node *curr)
{
while (curr != root && curr->color == black)
{//当前不为根，且为黑色
if (curr == curr->parent->left)
{//若其是父亲左孩子
node *brother = curr->parent->right;//兄弟节点肯定存在
if (brother->color == red)
{//情况1，兄弟是红色，转变为情况2,3,4
brother->color = black;
curr->parent->color = red;
leftRotate(curr->parent);
brother = curr->parent->right;
}
if (brother->left->color == black && brother->right->color == black)
{//情况2，兄弟是黑色，且两孩子也是黑色，将当前节点和兄弟去一重黑色
brother->color = red;
curr = curr->parent;
}
else if (brother->right->color == black)
{//情况3，兄弟左孩子为红，右孩子为黑，转变为情况4
brother->color = red;
brother->left->color = black;
rightRotate(brother);
brother = curr->parent->right;
}
else
{//情况4，右孩子为黑色，左孩子随意
brother->color = curr->parent->color;
curr->parent->color = black;
brother->right->color = black;
leftRotate(curr->parent);
curr = root;
}
}
else
{//若其是父亲右孩子，与上面四中情况对称
node *brother = curr->parent->left;
if (brother->color == red)
{//情况1
brother->color = black;
curr->parent->color = red;
rightRotate(curr->parent);
brother = curr->parent->left;
}
if (brother->right->color == black && brother->left->color == black)
{//情况2
brother->color = red;
curr = curr->parent;
}
else if (brother->left->color == black)
{//情况3
brother->color = red;
brother->right->color = black;
leftRotate(brother);
brother = curr->parent->left;
}
else
{//情况4
brother->color = curr->parent->color;
curr->parent->color = black;
brother->left->color = black;
rightRotate(curr->parent);
curr = root;
}
}
}
curr->color = black;//结束循环时将当前节点置为黑色
}
node* POMTree::locate(int k)const
{
node *curr = root;
while (curr != nil && curr->key != k)
{
if (k < curr->key)curr = curr->left;
else curr = curr->right;
}
return curr;
}
void POMTree::destroy()
{
while (root != nil)
{
//cout << "erase: " << root->key << endl;
erase(root->key);
}
delete nil;
}
int main()
{//16 21 8 9 25 30 5 8 15 23 19 20 17 19 26 26 0 3 6 10
//16 21 8 9 25 30 5 8 15 23 17 19 26 26 0 3 6 10 19 20
//7 9 5 8 0 3 6 10
POMTree ptree;
ptree.create();
cout << "inTraversal-----" << endl;
ptree.inTraversal();
cout << "preTraversal----" << endl;
ptree.preTraversal();
int low,high,choice;
cout << "Enter your choice,1-insert,2-delete,3-POM,4-print,5-exit" << endl;
do
{
cin >> choice;
switch(choice)
{
case 1:
cout << "Enter a interval,ex-> a b (a <= b): ";
cin >> low >> high;
ptree.insert(low,true);
ptree.insert(high,false);
break;
case 2:
cout << "Enter a interval,ex -> a b (a <= b): ";
cin >> low >> high;
ptree.erase(low);
ptree.erase(high);
break;
case 3:
ptree.findPom();
break;
case 4:
cout << "inTraversal-----" << endl;
ptree.inTraversal();
cout << "preTraversal----" << endl;
ptree.preTraversal();
break;
case 5:
ptree.destroy();
return 0;
default:
cout << "Bad choice" << endl;
}
//cout << "Enter your choice,1-insert,2-delete,3-POM,4-print,5-exit" << endl;
}while(choice >= 1 && choice <= 5);
ptree.destroy();
return 0;
}
对于main函数中的三组测试数据，细心的朋友可能已经发现第一组和第二组只有区间[19,20]的输入顺序不同，这也正是本代码存在的一个小bug，即如果存在这样的两个区间[x,y]和[y,z]，x <= y <= z,且y恰好是这些区间的最大重叠点，那么要保证得到正确结果，则需要将较靠后的区间先输入，即[y,z]先输入。这个bug我调了好久都没有找回解决方案，我估计还是更新或者初值设定有问题，如果哪位朋友知道是什么原因，希望能告知我，谢谢！
那么为了避免这个小bug，我们可以先将区间按左端点或者右端点从大到小排序，再输入；如果输入中不存在这样的区间，那最好了。
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