poj2580 Super Memmo

最新推荐文章于 2025-08-15 22:21:07 发布

原创最新推荐文章于 2025-08-15 22:21:07 发布 · 453 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#splay #poj #平衡树 #数据结构

平衡树同时被 2 个专栏收录

6 篇文章

订阅专栏

splay

4 篇文章

订阅专栏

本文解析了一道涉及Splay树的数据结构题目，通过实现多种操作如区间加值、翻转、旋转等，展示了Splay树的强大功能及其在解决复杂查询问题中的应用。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game. At first, the host tells the participant a sequence of numbers, {A₁, A₂, ... A_n}. Then the host performs a series of operations and queries on the sequence which consists:

ADD x y D: Add D to each number in sub-sequence {A_x ... A_y}. For example, performing "ADD 2 4 1" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5, 5}
REVERSE x y: reverse the sub-sequence {A_x ... A_y}. For example, performing "REVERSE 2 4" on {1, 2, 3, 4, 5} results in {1, 4, 3, 2, 5}
REVOLVE x y T: rotate sub-sequence {A_x ... A_y} T times. For example, performing "REVOLVE 2 4 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 2, 5}
INSERT x P: insert P after A_x. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
DELETE x: delete A_x. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
MIN x y: query the participant what is the minimum number in sub-sequence {A_x ... A_y}. For example, the correct answer to "MIN 2 4" on {1, 2, 3, 4, 5} is 2

To make the show more interesting, the participant is granted a chance to turn to someone else that means when Jackson feels difficult in answering a query he may call you for help. You task is to watch the TV show and write a program giving the correct answer to each query in order to assist Jackson whenever he calls.

Input

The first line contains n (n ≤ 100000).

The following n lines describe the sequence.

Then follows M (M ≤ 100000), the numbers of operations and queries.

The following M lines describe the operations and queries.

Output

For each "MIN" query, output the correct answer.

Sample Input

Sample Output

Source

POJ Founder Monthly Contest – 2008.04.13, Yao Jinyu
一道splay模板题。要求你设计一个数据结构实现以下操作：
1）Add(l,r,d)向区间l~r中的元素加上一个数d；
2）Reverse(l,r)翻转区间l~r的元素；
3）Revolve(l,r,T)将区间l~r的后缀放到区间的前方，操作T次；
4）Insert(x,P)将元素P插入到元素x后方；
5）Delete(x)删除元素x；
6）Min(l,r)求区间l~r中的最小值；
因为操作和题目需要，我们建立三个虚根，0、root（在NewNode操作后，root初始值为1）和root的右孩子（即ch[root][1]），一个实根，root的右孩子的左孩子（即ch[ch[root][1]][0]，Key_value的初始值）。

#include<cstdio>
#include<iostream>
#define INF 0x3f3f3f3f
#define Key_value ch[ch[root][1]][0]
using namespace std;
bool rev[200100];
int n,tot,root,a[200100],key[200100],add[200100],pre[200100],ch[200100][2],size[200100],minn[200100];
inline void Update_Add(int r,int d)
{
	if(!r)return;
	add[r]+=d;
	key[r]+=d;
	minn[r]+=d;
}
inline void Update_Reve(int r)
{
	if(!r)return;
	rev[r]^=1;
	swap(ch[r][0],ch[r][1]);
}
inline void Push_Up(int k)
{
	minn[k]=min(key[k],min(minn[ch[k][0]],minn[ch[k][1]]));
	size[k]=size[ch[k][0]]+size[ch[k][1]]+1;
}
inline void Push_Down(int r)
{
	if(add[r]){
		Update_Add(ch[r][0],add[r]);
		Update_Add(ch[r][1],add[r]);
		add[r]=0;
	}
	if(rev[r]){
		Update_Reve(ch[r][0]);
		Update_Reve(ch[r][1]);
		rev[r]=0;
	}
}
inline int Get_Kth(int r,int k)
{
	Push_Down(r);
	int t=size[ch[r][0]]+1;
	if(t==k)return r;
	if(t>k)return Get_Kth(ch[r][0],k);
	else return Get_Kth(ch[r][1],k-t);
}
inline void NewNode(int &r,int father,int val)
{
	r=++tot;
	size[r]=1;
	pre[r]=father;
	key[r]=minn[r]=val;
}
inline void Build(int &x,int father,int l,int r)
{
	if(l>r)return;
	int mid=(l+r)>>1;
	NewNode(x,father,a[mid]);
	Build(ch[x][0],x,l,mid-1);
	Build(ch[x][1],x,mid+1,r);
	Push_Up(x);
}
inline void Init()
{
	root=tot=0;
	minn[root]=INF;
	ch[root][0]=ch[root][1]=pre[root]=add[root]=rev[root]=size[root]=0;
	NewNode(root,0,INF);
	NewNode(ch[root][1],root,INF);
	Build(Key_value,ch[root][1],1,n);
}
inline void Rotate(int x,int kind)
{
	int y=pre[x];
	Push_Down(y);
	Push_Down(x);
	ch[y][!kind]=ch[x][kind];
	pre[ch[x][kind]]=y;
	if(pre[y])ch[pre[y]][ch[pre[y]][1]==y]=x;
	pre[x]=pre[y];
	ch[x][kind]=y;
	pre[y]=x;
	Push_Up(y);
}
inline void Splay(int r,int goal)
{
	Push_Down(r);
	while(pre[r]!=goal){
		if(pre[pre[r]]==goal){
			Push_Down(pre[r]);
			Push_Down(r);
			Rotate(r,ch[pre[r]][0]==r);
		}
		else{
			int y=pre[r];
			int kind=ch[pre[y]][0]==y;
			Push_Down(pre[y]);
			Push_Down(y);
			Push_Down(r);
			if(ch[y][kind]==r){
				Rotate(r,!kind);
				Rotate(r,kind);
			}
			else{
				Rotate(y,kind);
				Rotate(r,kind);
			}
		}
	}
	Push_Up(r);
	if(!goal)root=r;
}
inline void Add(int l,int r,int d)
{
	Splay(Get_Kth(root,l),0);//调用Get_Kth()首参数要用root因为root可能变成区间内其他数; 
	Splay(Get_Kth(root,r+2),root);
	Update_Add(Key_value,d);
	Push_Up(ch[root][1]);
	Push_Up(root);
}
inline int Min(int l,int r)
{
	Splay(Get_Kth(root,l),0);//ADD注释+1;
	Splay(Get_Kth(root,r+2),root);
	return minn[Key_value];
}
inline void Del(int x)
{
	Splay(Get_Kth(root,x),0);
	Splay(Get_Kth(root,x+2),root);
	pre[Key_value]=0;
	Key_value=0;//直接清零即可; 
	Push_Up(ch[root][1]);
	Push_Up(root);
}
inline void Reve(int l,int r)
{
	Splay(Get_Kth(root,l),0);
	Splay(Get_Kth(root,r+2),root);
	Update_Reve(Key_value);
	Push_Up(ch[root][1]);
	Push_Up(root);
}
inline void Revo(int a,int b,int t)
{
	int c=b-t;//将区间a~b分为两个区间a~b-c、b-c+1~b，将区间2旋转到区间1前;
	Splay(Get_Kth(root,a),0);
	Splay(Get_Kth(root,c+2),root);
	int tmp=Key_value;
	Key_value=0;
	Push_Up(ch[root][1]);
	Push_Up(root);
	Splay(Get_Kth(root,b-c+a),0);
	Splay(Get_Kth(root,b-c+a+1),root);
	Key_value=tmp;
	pre[Key_value]=ch[root][1];
	Push_Up(ch[root][1]);
	Push_Up(root);
}
inline void Ins(int x,int t)
{
	Splay(Get_Kth(root,x+1),0);
	Splay(Get_Kth(root,x+2),root);
	NewNode(Key_value,ch[root][1],t);
	Push_Up(ch[root][1]);
	Push_Up(root);
}
int main()
{
	while(scanf("%d",&n)!=EOF){
		for(int i=1;i<=n;i++)scanf("%d",&a[i]);
		Init();
		int Case;
		scanf("%d",&Case);
		for(;Case;Case--){
			char order[6];
			scanf("%s",order);
			if(order[0]=='A'){
				int x,y,z;
				scanf("%d%d%d",&x,&y,&z);
				Add(x,y,z);
			}
			if(order[0]=='M'){
				int x,y;
				scanf("%d%d",&x,&y);
				printf("%d\n",Min(x,y));
			}
			if(order[0]=='D'){
				int x;
				scanf("%d",&x);
				Del(x);
			}
			if(order[0]=='I'){
				int x,t;
				scanf("%d%d",&x,&t);
				Ins(x,t);
			}
			if(order[0]=='R'&&order[3]=='E'){
				int x,y;
				scanf("%d%d",&x,&y);
				Reve(x,y);
			}
			if(order[0]=='R'&&order[3]=='O'){
				int x,y,T;
				scanf("%d%d%d",&x,&y,&T);
				Revo(x,y,(T%(y-x+1)+y-x+1)%(y-x+1));
			}
		}
	}
	return 0;
}