POJ 3580 SuperMemo

裸Splay区间操作: 内存池+区间加减+区间翻转+插入+删除+维护最值


SuperMemo

Time Limit: 5000MS   Memory Limit: 65536K
Total Submissions: 8552   Accepted: 2801
Case Time Limit: 2000MS

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 ... AyT 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 (≤ 100000).

The following n lines describe the sequence.

Then follows 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

[Submit]   [Go Back]   [Status]  
[Discuss]

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=220000;
const int INF=0x3f3f3f3f;
#define Key_Value ch[ch[root][1]][0]

int ch[maxn][2],rev[maxn],add[maxn],sz[maxn],pre[maxn],key[maxn],minn[maxn];
int root,tot1;
int s[maxn],tot2;

int n,q,a[maxn];

void NewNode(int& x,int father,int k)
{
    if(tot2) x=s[tot2--];
    else x=++tot1;

    ch[x][0]=ch[x][1]=rev[x]=add[x]=0;
    sz[x]=1; pre[x]=father; key[x]=minn[x]=k;
}

void Erase(int r)
{
    if(r)
    {
        s[++tot2]=r;
        Erase(ch[r][0]);
        Erase(ch[r][1]);
    }
}

void Upd_Rev(int x)
{
    if(!x) return ;
    swap(ch[x][0],ch[x][1]);
    rev[x]^=1;
}

void Upd_Add(int x,int d)
{
    if(!x) return ;
    key[x]+=d; minn[x]+=d;
    add[x]+=d;
}

void Push_Up(int x)
{
    sz[x]=sz[ch[x][1]]+sz[ch[x][0]]+1;
    minn[x]=key[x];
    if(ch[x][0]) minn[x]=min(minn[x],minn[ch[x][0]]);
    if(ch[x][1]) minn[x]=min(minn[x],minn[ch[x][1]]);
}

void Push_Down(int x)
{
    if(rev[x])
    {
        Upd_Rev(ch[x][0]); Upd_Rev(ch[x][1]);
        rev[x]=0;
    }
    if(add[x])
    {
        Upd_Add(ch[x][0],add[x]); Upd_Add(ch[x][1],add[x]);
        add[x]=0;
    }
}

void Build(int& x,int l,int r,int fa)
{
    if(l>r) return ;
    int mid=(l+r)/2;
    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;
    ch[root][0]=ch[root][1]=pre[root]=sz[root]=0;
    minn[root]=key[root]=INF;
    NewNode(root,0,INF);
    NewNode(ch[root][1],root,INF);
    Build(Key_Value,1,n,ch[root][1]);
    Push_Up(ch[root][1]);
    Push_Up(root);
}

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];
    pre[y]=x;
    ch[x][kind]=y;
    Push_Up(y);
}

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);
            else Rotate(y,kind);
            Rotate(r,kind);
        }
    }
    Push_Up(r);
    if(goal==0) root=r;
}

int Get_Kth(int r,int k)
{
    Push_Down(r);
    int t=sz[ch[r][0]]+1;
    if(k==t) return r;
    if(t<k) return Get_Kth(ch[r][1],k-t);
    else return Get_Kth(ch[r][0],k);
}

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;
}

///.......do..it........

void ADD(int l,int r,int d)
{
    Splay(Get_Kth(root,l),0);
    Splay(Get_Kth(root,r+2),root);
    Upd_Add(Key_Value,d);
    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);
    Upd_Rev(Key_Value);
    Push_Up(ch[root][1]);
    Push_Up(root);
}

void DELETE(int p)
{
    Splay(Get_Kth(root,p),0);
    Splay(Get_Kth(root,p+2),root);
    Erase(Key_Value);
    pre[Key_Value]=0;
    Key_Value=0;
    Push_Up(ch[root][1]);
    Push_Up(root);
}

void INSERT(int p,int v)
{
    Splay(Get_Kth(root,p+1),0);
    Splay(Get_Kth(root,p+2),root);
    NewNode(Key_Value,ch[root][1],v);
    Push_Up(ch[root][1]);
    Push_Up(root);
}

int MIN(int l,int r)
{
    Splay(Get_Kth(root,l),0);
    Splay(Get_Kth(root,r+2),root);
    return minn[Key_Value];
}

void REVOLVE(int l,int r,int t)
{
    int len=r-l+1;
    t=(t%len+len)%len;
    if(t==0) return ;
    int c=r-t+1;
    ///[l..c-1] and [c..r]
    Splay(Get_Kth(root,c),0);
    Splay(Get_Kth(root,r+2),root);
    int temp=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=temp;
    pre[Key_Value]=ch[root][1];
    Push_Up(ch[root][1]);
    Push_Up(root);
}
/****************Debug*********************/

void showit(int x)
{
    if(x)
    {
        Push_Down(x);
        showit(ch[x][0]);
        printf("结点: %2d key: %2d 左儿子: %2d 右儿子: %2d 父结点: %2d sz: %2d min: %2d \n",
               x,key[x],ch[x][0],ch[x][1],pre[x],sz[x],minn[x]);
        showit(ch[x][1]);
    }
}

void Debug()
{
    cout<<"---------------------------\n";
    cout<<"root: "<<root<<endl;
    showit(root);
}

int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=1;i<=n;i++) scanf("%d",a+i);
        Init();
       // Debug();
        scanf("%d",&q);
        char op[100]; int x,y,z;
        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 if(strcmp(op,"MIN")==0)
            {
                scanf("%d%d",&x,&y);
                printf("%d\n",MIN(x,y));
            }
           // Debug();
        }
    }
    return 0;
}

POJ 3580 SuperMemo,布布扣,bubuko.com

时间: 2024-08-25 03:26:02

POJ 3580 SuperMemo的相关文章

poj 3580 SuperMemo (Splay)

poj 3580 好恶心的题目,真是各种操作都来了个遍 ... 不过Splay树真是一种神奇的东西,通过旋转就能实现各种操作,而且方法也都相差不大 . 题意: 给出一个数字序列,有6种操作: (1) ADD x y d: 第x个数到第y个数加d . (2) REVERSE x y : 将区间[x,y]中的数翻转 . (3) REVOLVE x y t :将区间[x,y]旋转t次,如1 2 3 4 5 旋转2次后就变成4 5 1 2 3 . (4) INSERT x p :在第x个数后面插入p .

poj 3580 SuperMemo splay树模板题

题意: 给一个序列,对其进行各种操作.在对序列仅对操作的处理上,splay是比线段树强大的,虽然均摊复杂度均为logN,但它能支持1:在某个位置插入一些连续的数,2:在某个位置删除一些连续的数.只是splay随便一些200+行. 分析: 网上各种模板介绍漫天飞,这个还算简洁明了. 代码: //poj 3580 #include <stdio.h> #define maxN 200000 int N,T,node; int a[maxN],size[maxN],left[maxN],right[

POJ 3580 - SuperMemo - [伸展树splay]

题目链接:http://poj.org/problem?id=3580 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 h

Splay树(多操作)——POJ 3580 SuperMemo

对应POJ题目:点击打开链接 SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 11309   Accepted: 3545 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a

又一道Splay吐血题 [POJ 3580] SuperMemo

SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 9878   Accepted: 3177 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game.

POJ 3580 SuperMemo (Splay 区间更新、翻转、循环右移,插入,删除,查询)

SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 13917   Accepted: 4352 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game

平衡树:Splaytree POJ 3580 SuperMemo

SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 12788   Accepted: 3986 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which the participant is told to play a memorizing game

POJ 3580 SuperMemo(伸展树的基本操作)

题目大意:给你六个操作,让你实现这些功能. 解题思路:伸展树的基本应用,用伸展数实现各种功能. SuperMemo Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 10404   Accepted: 3320 Case Time Limit: 2000MS Description Your friend, Jackson is invited to a TV show called SuperMemo in which t

POJ 3580 SuperMemo 题解

题目大意:维护一个序列,支持6种操作: 1.ADD x y D 从第x个数到第y个数都增加D 2.REVERSE x y 翻转第x个数到第y个数 3.REVOLVE x y T 从x到y,向右循环移动T次 4.INSERT x P 插入P到第x个数后面 5.DELETE x 删除第x个数 6.MIN x y 查询第x个数到第y个数之间最小值 ADD和REVERSE打标记然后及时下传就可以..REVOLVE的T可能为负还可能很大,要注意取模.REVOLVE可以通过3个REVERSE操作完成,也可以