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| 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. 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:
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
题意：给你一个序列以及许多操作，包括区间更新、翻转、循环右移，插入，删除，查询
分析：都是Splay的一些基本的操作，更新和翻转使用lazy标记的思想就行了，对于循环右移，首先找出左端点移动的位置k，然后就是将[k,r]区间移动到[l,k-1]的前面去，我们可以先定位出[k,r]，再调整出
[l-1,l],将[k,r]接上去就行了。插入操作也是通过变换找到这个子树，再新加节点就是了。
#include<stdio.h> #include<string.h> #include<iostream> using namespace std; #define Key_value ch[ch[root][1]][0] const int MAXN = 200000+100; const int INF = 0X3F3F3F3F; int fa[MAXN],ch[MAXN][2],key[MAXN],sz[MAXN],add[MAXN],rev[MAXN],mi[MAXN]; int root,tot1; int s[MAXN],tot2;//内存池、内存池容量 int a[MAXN]; int n,m; void NewNode(int &r,int pre,int k) { if(tot2) r=s[tot2--]; else r=++tot1; ch[r][0]=ch[r][1]=0; fa[r]=pre; sz[r]=1; add[r]=rev[r]=0; key[r]=mi[r]=k; } void Update_Rev(int r) { if(r==0) return; swap(ch[r][0],ch[r][1]); rev[r]^=1; } void Update_Add(int r,int val) { if(r==0) return; add[r]+=val; key[r]+=val; mi[r]+=val; } void Push_Up(int r) { sz[r]=sz[ch[r][0]]+sz[ch[r][1]]+1; mi[r]=key[r]; if(ch[r][0]) mi[r]=min(mi[r],mi[ch[r][0]]); if(ch[r][1]) mi[r]=min(mi[r],mi[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 pre) { if(l>r) return; int mid=(l+r)/2; NewNode(x,pre,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]=sz[root]=add[root]=rev[root]=fa[root]=0; mi[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 d) { int y=fa[x]; Push_Down(y); Push_Down(x); ch[y][!d]=ch[x][d]; fa[ch[x][d]]=y; if(fa[y]) ch[fa[y]][ch[fa[y]][1]==y]=x; fa[x]=fa[y]; ch[x][d]=y; fa[y]=x; Push_Up(y); } //伸展操作，将r调整到goal下面 void Splay(int r,int goal) { Push_Down(r); while(fa[r]!=goal) { if(fa[fa[r]]==goal) { //这题有反转操作，需要先push_down,在判断左右孩子 Push_Down(fa[r]); Push_Down(r); Rotate(r,ch[fa[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(fa[fa[r]]); Push_Down(fa[r]); Push_Down(r); int y=fa[r]; int d=(ch[fa[y]][0]==y); //两个方向不同，则先左旋再右旋 if(ch[y][d]==r) { Rotate(r,!d); Rotate(r,d); } //两个方向相同，相同方向连续两次 else { Rotate(y,d); Rotate(r,d); } } } 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(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; } void Erase(int r) { if(r) { s[++tot2]=r; Erase(ch[r][0]); Erase(ch[r][1]); } } int Get_Max(int r) { Push_Down(r); while(ch[r][1]) { r=ch[r][1]; Push_Down(r); } return r; } void update_add(int l,int r,int val) { Splay(Get_Kth(root,l),0); Splay(Get_Kth(root,r+2),root); Update_Add(Key_value,val); Push_Up(ch[root][1]); Push_Up(root); } void update_rev(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 update_reve(int l,int r,int T) { int len=r-l+1; T=(T%len+len)%len; if(T==0) return; int k=r-T+1; Splay(Get_Kth(root,k),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; fa[Key_value]=ch[root][1]; Push_Up(ch[root][1]); Push_Up(root); } void update_int(int x,int 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 update_del(int x) { Splay(Get_Kth(root,x),0); Splay(Get_Kth(root,x+2),root); Erase(Key_value); fa[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 mi[Key_value]; } char str[20]; int x,y,D,P,T; int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%d",&a[i]); Init(); scanf("%d",&m); while(m--) { scanf("%s",str); if(strcmp(str,"ADD")==0) { scanf("%d%d%d",&x,&y,&D); update_add(x,y,D); } else if(strcmp(str,"REVERSE")==0) { scanf("%d%d",&x,&y); update_rev(x,y); } else if(strcmp(str,"REVOLVE")==0) { scanf("%d%d%d",&x,&y,&T); update_reve(x,y,T); } else if(strcmp(str,"INSERT")==0) { scanf("%d%d",&x,&P); update_int(x,P); } else if(strcmp(str,"DELETE")==0) { scanf("%d",&x); update_del(x); } else { scanf("%d%d",&x,&y); printf("%d\n",query_min(x,y)); } } return 0; }
POJ 3580 SuperMemo (Splay 区间更新、翻转、循环右移，插入，删除，查询)
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原文地址：http://www.cnblogs.com/wangdongkai/p/5755144.html
