SuperMemo
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, ... An}. 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 {Ax ... Ay}. 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 {Ax ... Ay}. 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 {Ax ... Ay} 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 Ax. For example, performing "INSERT 2 4" on {1, 2, 3, 4, 5} results in {1, 2, 4, 3, 4, 5}
- DELETE x: delete Ax. 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 {Ax ... Ay}. 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
5 1 2 3 4 5 2 ADD 2 4 1 MIN 4 5
Sample Output
5
这篇博客用来保存伸展树模板！……看了好久的伸展树，也只是似懂非懂，但至少要会用。因此总结了一个模板。除了题目包含的操作外，还新增了几个操作。详看代码！
#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string.h>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#include<bitset>
#define Key_Value ch[ch[root][1]][0]
using namespace std;
typedef long long int ll;
const int MAXN=300010;
const int INF=0x3f3f3f3f;
//伸展树模板！
int pre[MAXN];//父节点
int ch[MAXN][2];//左右两个孩子
int size[MAXN];//子树大小
int root;//根节点
int tot1;//总大小
int key[MAXN];//该点的值
long long int sum[MAXN];//子树和
int minnum[MAXN];//最小值
int lazySum[MAXN];//区间累加延迟标记
int lazyRev[MAXN];//翻转区间延迟标记
int lazyCha[MAXN];//区间修改延迟标记
int s[MAXN],tot2;//用于删除操作
int A[MAXN];//原始数组
int n,q;
//新建节点，节点编号，父亲编号，值
void NewNode(int &r,int fa,int k){
if(tot2)
r=s[tot2--];
else
r=++tot1;
pre[r]=fa;
size[r]=1;
key[r]=k;
lazyRev[r]=0;
minnum[r]=k;
lazySum[r]=0;
lazyCha[r]=0;
sum[r]=0;
ch[r][0]=ch[r][1]=0;
}
//给r为根的子树增加值
void Update_Add(int r,int C){
if(r==0)
return;
lazySum[r]+=C;
key[r]+=C;
minnum[r]+=C;
sum[r]+=(long long)C*size[r];
}
void Update_Same(int r,int v)
{
if(!r)return;
minnum[r]=v;
key[r]=v;
sum[r]=v*size[r];
lazyCha[r]=1;
}
void Update_Rev(int r){
if(!r)
return;
swap(ch[r][0],ch[r][1]);
lazyRev[r]^=1;
}
void Push_Up(int r){
size[r]=size[ch[r][0]]+size[ch[r][1]]+1;
sum[r]=sum[ch[r][0]]+sum[ch[r][1]]+key[r];
minnum[r]=key[r];
if(ch[r][0])minnum[r]=min(minnum[r],minnum[ch[r][0]]);
if(ch[r][1])minnum[r]=min(minnum[r],minnum[ch[r][1]]);
}
void Push_Down(int r){
if(lazyRev[r]){
Update_Rev(ch[r][0]);
Update_Rev(ch[r][1]);
lazyRev[r]=0;
}
if(lazySum[r]){
Update_Add(ch[r][0],lazySum[r]);
Update_Add(ch[r][1],lazySum[r]);
lazySum[r]=0;
}
if(lazyCha[r])
{
Update_Same(ch[r][0],key[r]);
Update_Same(ch[r][1],key[r]);
lazyCha[r]=0;
}
}
//建树
void Build(int &x,int l,int r,int fa){
if(l>r)
return;
int mid=(l+r)>>1;
NewNode(x,fa,A[mid]);
Build(ch[x][0],l,mid-1,x);
Build(ch[x][1],mid+1,r,x);
Push_Up(x);
}
void Init(){
root=tot1=tot2=0;
lazyCha[root]=ch[root][0]=ch[root][1]=pre[root]=size[root]=lazySum[root]=lazyRev[root]=sum[root]=0;
key[root]=0;
minnum[root]=INF;
// NewNode(root,0,INF);
// NewNode(ch[root][1],root,INF);//求最小值用
NewNode(root,0,-1);
NewNode(ch[root][1],root,-1);
Build(Key_Value,1,n,ch[root][1]);
Push_Up(ch[root][1]);
Push_Up(root);
}
//0左旋，1右旋
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);
}
//伸展，将r调到goal下
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{
Push_Down(pre[pre[r]]);
Push_Down(pre[r]);
Push_Down(r);
int y=pre[r];
int kind=ch[pre[y]][0]==y;
if(ch[y][kind]==r){
Rotate(r,!kind);
Rotate(r,kind);
}
else{
Rotate(y,kind);
Rotate(r,kind);
}
}
}
Push_Up(r);
if(goal==0)
root=r;
}
void erase(int r)//回收内存
{
if(r)
{
s[++tot2]=r;
erase(ch[r][0]);
erase(ch[r][1]);
}
}
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);
}
int Get_Min(int r){
Push_Down(r);
while(ch[r][0]){
r=ch[r][0];
Push_Down(r);
}
return r;
}
int Get_Max(int r){
Push_Down(r);
while(ch[r][1]){
r=ch[r][1];
Push_Down(r);
}
return r;
}
void Add(int l,int r,int C){
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,r+2),root);
Update_Add(ch[ch[root][1]][0],C);
Push_Up(ch[root][1]);
Push_Up(root);
}
void Reverse(int l,int r)
{
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,r+2),root);
Update_Rev(Key_Value);
Push_Up(ch[root][1]);
Push_Up(root);
}
void Revolve(int l,int r,int T)//循环右移
{
int len=r-l+1;
T=(T%len+len)%len;
if(T==0)return;
int c=r-T+1;//将区间[c,r]放在[l,c-1]前面
Splay(Get_Kth(root,c),0);
Splay(Get_Kth(root,r+2),root);
int tmp=Key_Value;
Key_Value=0;
Push_Up(ch[root][1]);
Push_Up(root);
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,l+1),root);
Key_Value=tmp;
pre[Key_Value]=ch[root][1];//这个不用忘记
Push_Up(ch[root][1]);
Push_Up(root);
}
void Insert(int x,int P)//在第x个数后面插入P
{
Splay(Get_Kth(root,x+1),0);
Splay(Get_Kth(root,x+2),root);
NewNode(Key_Value,ch[root][1],P);
Push_Up(ch[root][1]);
Push_Up(root);
}
//在第pos个数后插入tot个数
void Insert_Range(int pos,int tot)
{
for(int i=0;i<tot;i++)
scanf("%d",&A[i]);
Splay(Get_Kth(root,pos+1),0);
Splay(Get_Kth(root,pos+2),root);
Build(Key_Value,0,tot-1,ch[root][1]);
Push_Up(ch[root][1]);
Push_Up(root);
}
long long int Query_Sum(int l,int r){
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,r+2),root);
return sum[ch[ch[root][1]][0]];
}
void Delete(int x)//删除第x个数
{
Splay(Get_Kth(root,x),0);
Splay(Get_Kth(root,x+2),root);
erase(Key_Value);
pre[Key_Value]=0;
Key_Value=0;
Push_Up(ch[root][1]);
Push_Up(root);
}
//从第pos个数开始连续删除tot个数
void Delete(int pos,int tot)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
erase(Key_Value);
pre[Key_Value]=0;
Key_Value=0;
Push_Up(ch[root][1]);
Push_Up(root);
}
//从第pos个数连续开始的tot个数修改为c
void Make_Same(int pos,int tot,int c)
{
Splay(Get_Kth(root,pos),0);
Splay(Get_Kth(root,pos+tot+1),root);
Update_Same(Key_Value,c);
Push_Up(ch[root][1]);
Push_Up(root);
}
int Query_Min(int l,int r)
{
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,r+2),root);
return minnum[Key_Value];
}
//将l,r区间搬到c后
void Move(int l,int r,int c)
{
Splay(Get_Kth(root,l),0);
Splay(Get_Kth(root,r+2),root);
int tmp=Key_Value;
Key_Value=0;
Push_Up(ch[root][1]);
Push_Up(root);
Splay(Get_Kth(root,c+1),0);
Splay(Get_Kth(root,c+2),root);
Key_Value=tmp;
pre[Key_Value]=ch[root][1];
Push_Up(ch[root][1]);
Push_Up(root);
}
int cnt=0;
void Print(int r=root)
{
if(!r)return;
Push_Down(r);
Print(ch[r][0]);
if(cnt>=1&&cnt<=n)
{
printf("%d",key[r]);
if(cnt<n)printf(" ");
else printf("\n");
}
cnt++;
Print(ch[r][1]);
}
int main(){
char op[20];
int x,y,z;
while(scanf("%d",&n)==1)
{
for(int i=1;i<=n;i++)scanf("%d",&A[i]);
Init();
scanf("%d",&q);
while(q--)
{
scanf("%s",op);
if(strcmp(op,"ADD")==0)
{
scanf("%d%d%d",&x,&y,&z);
Add(x,y,z);
}
else if(strcmp(op,"REVERSE")==0)
{
scanf("%d%d",&x,&y);
Reverse(x,y);
}
else if(strcmp(op,"REVOLVE")==0)
{
scanf("%d%d%d",&x,&y,&z);
Revolve(x,y,z);
}
else if(strcmp(op,"INSERT")==0)
{
scanf("%d%d",&x,&y);
Insert(x,y);
}
else if(strcmp(op,"DELETE")==0)
{
scanf("%d",&x);
Delete(x);
}
else
{
scanf("%d%d",&x,&y);
printf("%d\n",Query_Min(x,y));
}
}
}
return 0;
}
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