POJ 3580 SuperMemo（splay成段更新、区间最小值、反转、插入和删除、区间搬移）

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POJ 3580 SuperMemo（成段更新、区间最小值、反转、插入和删除、区间搬移）

SuperMemo

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 5839		Accepted: 1884
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, ... 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

5都是伸展树比较经典的操作。
特别的是其中的循环右移操作。循环右移[l,r] T次，其实就是把区间[l,r-T]放在[r-T+1,r]后面。就是区间搬移。但是T必须先对长度取模

狂斌大神模板

/** 给定一个数列a1,a2,...an* 进行以下6种操作* ADD x y D :给第x个数到第y个数加D(增加一个add进行延迟标记)* REVERSE x y :反转[x,y]之间的数(伸展树经典操作)* REVOLVE x y T:循环右移T次(先把T对长度进行取模，然后就相当于把[y-T+1,y]放在[x,y-T]前面)* INSERT x P:在第x个数后面插入P （经典的插入）* DELETE x:删除第x个数（删除操作）* MIN x y:查询[x,y]之间最小的数(标记)** 需要的操作:反转、删除、插入、查询区间最小值、成段更新、区间搬移(循环右移转化为区间搬移)* 需要的变量:pre,ch,key,size,add,rev,m(最小值)*/#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
#define Key_value ch[ch[root][1]][0]
#define Key_Value ch[ch[root][1]][0]
const int MAXN=200010;
const int INF=0x3f3f3f3f;
int pre[MAXN],ch[MAXN][2],key[MAXN],size[MAXN],add[MAXN],rev[MAXN],m[MAXN];
int root,tot1;
int s[MAXN],tot2;//内存池、内存池容量
int a[MAXN];
int n,q;void NewNode(int &r,int father,int k)
{if(tot2)r=s[tot2--];else r=++tot1;ch[r][0]=ch[r][1]=0;pre[r]=father;size[r]=1;add[r]=rev[r]=0;key[r]=m[r]=k;
}
void Update_Rev(int r)
{if(!r)return;swap(ch[r][0],ch[r][1]);rev[r]^=1;
}
void Update_Add(int r,int ADD)
{if(!r)return;add[r]+=ADD;key[r]+=ADD;m[r]+=ADD;
}
void Push_Up(int r)
{size[r]=size[ch[r][0]]+size[ch[r][1]]+1;m[r]=key[r];if(ch[r][0])m[r]=min(m[r],m[ch[r][0]]);if(ch[r][1])m[r]=min(m[r],m[ch[r][1]]);
}
void Push_Down(int r)
{if(rev[r]){Update_Rev(ch[r][0]);Update_Rev(ch[r][1]);rev[r]=0;}if(add[r]){Update_Add(ch[r][0],add[r]);Update_Add(ch[r][1],add[r]);add[r]=0;}
}
void Build(int &x,int l,int r,int father)
{if(l>r)return;int mid=(l+r)/2;NewNode(x,father,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]=size[root]=add[root]=rev[root]=pre[root]=0;m[root]=INF;//这个不用也可以，如果在push_up那判断了的话，否则需要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);
}
//旋转
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);
}
//Splay调整
void Splay(int r,int goal)
{Push_Down(r);while(pre[r]!=goal){if(pre[pre[r]]==goal){//这题有反转操作，需要先push_down,在判断左右孩子Push_Down(pre[r]);Push_Down(r);Rotate(r,ch[pre[r]][0]==r);}else{//这题有反转操作，需要先push_down,在判断左右孩子Push_Down(pre[pre[r]]);Push_Down(pre[r]);Push_Down(r);int y=pre[r];int kind=(ch[pre[y]][0]==y);//两个方向不同，则先左旋再右旋if(ch[y][kind]==r){Rotate(r,!kind);Rotate(r,kind);}//两个方向相同，相同方向连续两次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=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;
}
//下面是操作了
void ADD(int l,int r,int D)
{Splay(Get_Kth(root,l),0);Splay(Get_Kth(root,r+2),root);Update_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);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);
}
void erase(int r)//回收内存
{if(r){s[++tot2]=r;erase(ch[r][0]);erase(ch[r][1]);}
}
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);
}
int Query_Min(int l,int r)
{Splay(Get_Kth(root,l),0);Splay(Get_Kth(root,r+2),root);return m[Key_value];
}
int main()
{//freopen("in.txt","r",stdin);//freopen("out.txt","w",stdout);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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