POJ 3580 SuperMemo（splay成段更新、区间最小值、反转、插入和删除、区间搬移）
来源：互联网 发布：对淘宝客服的理解 编辑：程序博客网 时间：2024/06/06 12:50
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: 65536KTotal Submissions: 5839
Accepted: 1884Case Time Limit: 2000MS
Memory Limit: 65536KTotal Submissions: 5839
Accepted: 1884Case 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:
- 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
51 2 3 4 52ADD 2 4 1MIN 4 5Sample 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;}
0 0
- POJ 3580 SuperMemo（splay成段更新、区间最小值、反转、插入和删除、区间搬移）
- POJ 3580 SuperMemo (Sply经典操作:成段更新、区间最小值、反转、插入和删除、区间搬移)
- POJ 题目3580 SuperMemo（Splay Tree区间加，区间翻转，区间右移，插入删除，区间最小值）
- POJ 3580 SuperMemo Splay 区间维护
- POJ 3580 SuperMemo (Splay各种区间操作)
- POJ 3580(SuperMemo-Splay区间加)[template:Splay V2]
- 伸展树（区间加值，反转，循环移动，插入，删除区间，求区间最小值）poj3580
- POJ 3580 SuperMemo Splay树各种区间操作
- POJ 3580 SuperMemo Splay/可持久化Treap 维护区间
- 【数据结构SPLAY】splay区间翻转，区间和查询，区间插入，区间删除，区间修改
- POJ 3468 A Simple Problem with Integers （splay 成段更新、区间求和）
- POJ 3580 SuperMemo （Splay）
- 新技能：splay支持区间翻转，区间插入，区间删除
- Poj 3667 Hotel 成段更新，区间合并，区间求和
- Splay 区间反转
- splay POJ 3580SuperMemo
- POJ 3580 SuperMemo(Splay)
- POJ 3580 SuperMemo(Splay)
- 史上最详尽的平衡树(splay)讲解与模板
- redis持久化
- android 串口通讯 JNI
- Codeforces 777E/778C 题解 （贪心）
- Python 爬虫学习4
- POJ 3580 SuperMemo（splay成段更新、区间最小值、反转、插入和删除、区间搬移）
- 使用Masonry还是storyboard？
- 看门狗框架的原理
- Winform应用程序实现通用遮罩层二
- JSP标准标签库——JSTL
- 【LeetCode题解】56. Merge Intervals
- C++Primer第五版 第十三章习题答案（21~30）
- C/S架构的简单文件传输系统的实现
- 芬语录3.6
