poj2580 Super Memmo

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:

  1. 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}
  2. 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}
  3. 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}
  4. 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}
  5. DELETE x: delete Ax. For example, performing "DELETE 2" on {1, 2, 3, 4, 5} results in {1, 3, 4, 5}
  6. 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

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
",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;
}
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原文地址：https://www.cnblogs.com/keshuqi/p/5957774.html