7
12
2019
2

BZOJ 1588: [HNOI2002]营业额统计 && 红黑树板子

Time Limit: 5 Sec  Memory Limit: 162 MB

Description

营业额统计 Tiger最近被公司升任为营业部经理,他上任后接受公司交给的第一项任务便是统计并分析公司成立以来的营业情况。 Tiger拿出了公司的账本,账本上记录了公司成立以来每天的营业额。分析营业情况是一项相当复杂的工作。由于节假日,大减价或者是其他情况的时候,营业额会出现一定的波动,当然一定的波动是能够接受的,但是在某些时候营业额突变得很高或是很低,这就证明公司此时的经营状况出现了问题。经济管理学上定义了一种最小波动值来衡量这种情况: 该天的最小波动值 当最小波动值越大时,就说明营业情况越不稳定。 而分析整个公司的从成立到现在营业情况是否稳定,只需要把每一天的最小波动值加起来就可以了。你的任务就是编写一个程序帮助Tiger来计算这一个值。 第一天的最小波动值为第一天的营业额。 

Input

第一行为正整数 ,表示该公司从成立一直到现在的天数,接下来的n行每行有一个整数(有可能有负数) ,表示第i天公司的营业额。

Output

输出文件仅有一个正整数,即Sigma(每天最小的波动值) 。结果小于2^31 。

Sample Input

6
5
1
2
5
4
6

Sample Output

12

HINT

结果说明:5+|1-5|+|2-1|+|5-5|+|4-5|+|6-5|=5+4+1+0+1+1=12

继续学习红黑树的结点删除QaQ

#include 
#include 

using namespace std;

struct node
{
	bool Color; //True: Red  False: Black
	int Key;
	node* p;
	node* left, * right;
};

int ans = 0;
node* Root;//根节点
node* NIL;//空节点

#ifdef _DEBUG
template
constexpr auto min(Type a, Type b) { if (a < b)return a; else return b; }
#else
#define min(a,b) ((a)<(b)?(a):(b))
#endif // _DEBUG

void Init()
{
	NIL = new node;					//定义空哨兵节点(叶子结点和根节点的父节点)
	NIL->Color = false;
	NIL->left = NIL->right = NULL;
	NIL->Key = 0;

	Root = NULL;
}

int Insert(node* pnt, node *z)		//插入元素 返回值为与树内结点差绝对值最小值
{
	if (pnt == NULL)				//空红黑树 p为根节点
	{
		Root = z;
		z->p = NIL;
		return z->Key;
	}
	int temp;//临时变量 用于记录下一层Insert返回值
	if (z->Key < pnt->Key)//z将被加入左子树豪华午餐
	{
		if (pnt->left == NIL)
		{
			pnt->left = z;
			z->p = pnt;
			return pnt->Key - z->Key;
		}
		else
		{
			temp = Insert(pnt->left, z);
			return min(temp, pnt->Key - z->Key);
		}
	}
	else //z将被加入右子树奢华晚餐
	{
		if (pnt->right == NIL)
		{
			pnt->right = z;
			z->p = pnt;
			return z->Key - pnt->Key;
		}
		else
		{
			temp = Insert(pnt->right, z);
			return min(temp, z->Key - pnt->Key);
		}
	}
}

void Shift_L(node* pnt)//树左旋
{
	node* temp = pnt->right;
	pnt->right = temp->left;
	temp->left->p = pnt;
	temp->p = pnt->p;
	if (pnt->p->left == pnt)pnt->p->left = temp;
	else pnt->p->right = temp;
	pnt->p = temp;
	temp->left = pnt;
	if (pnt == Root)Root = temp;
}

void Shift_R(node* pnt)//树右旋
{
	node* temp = pnt->left;
	pnt->left = temp->right;
	temp->right->p = pnt;
	temp->p = pnt->p;
	if (pnt->p->left == pnt)pnt->p->left = temp;
	else pnt->p->right = temp;
	pnt->p = temp;
	temp->right = pnt;
	if (pnt == Root)Root = temp;
}

void Fix_up(node* z)
{
	while (z->p->Color)//z的父节点为红色(z也为红色)不符合红黑树定义
	{
		if (z->p->p->left == z->p)//z的父节点在祖父节点的左子树上
		{
			if (z->p->p->right->Color)//叔节点为红色
			{
				z->p->p->Color = true;
				z->p->p->left->Color = z->p->p->right->Color = false;
				z = z->p->p;
			}
			else
			{
				if (z == z->p->right)//z在右节点上,左旋一下
				{
					z = z->p;
					Shift_L(z);
				}
				z->p->Color = false;
				z->p->p->Color = true;
				Shift_R(z->p->p);
			}
		}
		else //z的父节点在z的祖父节点右子树上 所有操作左右相反即可
		{
			if (z->p->p->left->Color)//叔节点为红色
			{
				z->p->p->Color = true;
				z->p->p->left->Color = z->p->p->right->Color = false;
				z = z->p->p;
			}
			else
			{
				if (z == z->p->left)//z在左节点上,右旋一下
				{
					z = z->p;
					Shift_R(z);
				}
				z->p->Color = false;
				z->p->p->Color = true;
				Shift_L(z->p->p);
			}
		}
	}
	Root->Color = false;
}

void TreeOut(node* Pos)
{
	if (Pos == NIL)return;
	printf("%d %d l:%d r:%d p:%d\n", Pos->Key, Pos->Color, Pos->left->Key, Pos->right->Key, Pos->p->Key);
	TreeOut(Pos->left);
	TreeOut(Pos->right);
}

int n, x;
int main()
{
	Init();
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i)
	{
		scanf("%d", &x);
		node* z = new node;
		z->Key = x;
		z->left = z->right = NIL;
		z->Color = true;
		ans += Insert(Root, z);
		Fix_up(z);
#ifdef _DEBUG
		printf("-------------%d------------\n", x);
		TreeOut(Root);
#endif // _DEBUG
	}
	printf("%d\n", ans);
	return 0;
}
Category: 程序 | Tags: 红黑树 bzoj | Read Count: 504
Avatar_small
Assam HSLC Question 说:
2022年8月27日 15:18

Every year Students Struggle to acquire the Pass marks in 10th Examination. Student Download Assam HSLC Question Paper 2023 After Regular Prepare to Participate in must take a printout of Assam Board HSLC Model Paper 2023 available our Website. Student’s use Previous Paper and Start Preparation Assam HSLC Final Exam 2023 Every year Students Struggle to acquire the Pass marks in 10th Examination. Assam HSLC Question Paper Student Download Assam HSLC Question Paper 2023 After Regular Prepare to Participate in must take a printout of Assam Board HSLC Model Paper 2023 available our Website. Student’s use Previous Paper and Start Preparation Assam HSLC Final Exam 2023

Avatar_small
AP 1st Inter Botany 说:
2022年9月08日 19:34

In intermediate education, Botany is a very important subject for science group students for part-1 and part-2 first and the second year of junior and senior intermediate students, and the BIEAP has conducted the General Science examination tests including botany subject, download AP inter Botany Model Paper 2023 AP 1st Inter Botany Model Paper Pdf with solutions for all 1st and 2nd-year students for paper-1 and paper-2 exams 2023.


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter

Host by is-Programmer.com | Power by Chito 1.3.3 beta | Theme: Aeros 2.0 by TheBuckmaker.com