【poj3580】 SuperMemo

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:

1. 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}
2. 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}
3. 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}
4. 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}
5. DELETE x: delete A_x. 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 {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

51 2 3 4 52ADD 2 4 1MIN 4 5

Sample Output

5

初学splay感觉身体被掏空 辣鸡splay 毁我青春

这是splay的模板题，但是非常BT，所以磨了将近10个小时才搞定。

参考学习了大神cxlove的写法orzorzorz%。网址：http://blog.csdn.net/acm_cxlove/article/details/7803646

splay+线段树的结构坑死人，记得处处都要pushup和pushdown。而且最好不要手残（像我）。找问题会找到疯。

revolve操作比较坑，可以理解为区间[a,b]经过转换变成区间[c,b]+[a,c-1],c大小可以通过计算得出。

剩下看代码吧。

#include<iostream>#include<cstdio>#include<cstring>#include<cstdlib>#include<algorithm>#define key_root ch[ch[root][1]][0]#define M 400050#define inf 2000000000using namespace std;int pre[M],size[M],rev[M],key[M];int ch[M][2],m[M],add[M];int a[M];int tot,root,n;void pr(int r){if(r==0)return;pr(ch[r][0]);cout<<r<<" "<<key[r]<<" "<<ch[r][0]<<" "<<ch[r][1]<<" "<<pre[r]<<endl;pr(ch[r][1]);}inline void update_add(int r,int c){if(!r)return;add[r]+=c;key[r]+=c;m[r]+=c;}inline void update_rev(int r){if(!r)return;rev[r]^=1;swap(ch[r][0],ch[r][1]);}inline void pushup(int x){size[x]=size[ch[x][0]]+size[ch[x][1]]+1;m[x]=min(key[x],min(m[ch[x][0]],m[ch[x][1]]));}inline void pushdown(int x){if(add[x]){update_add(ch[x][0],add[x]);update_add(ch[x][1],add[x]);add[x]=0;}if(rev[x]){update_rev(ch[x][0]);update_rev(ch[x][1]);rev[x]=0;}}inline void newnode(int &r,int father,int a){r=++tot;ch[r][0]=ch[r][1]=0;pre[r]=father;key[r]=m[r]=a;size[r]=1;add[r]=rev[r]=0;}void build(int &x,int fa,int l,int r){if(l>r)return;int mid=(l+r)>>1;newnode(x,fa,a[mid]);build(ch[x][0],x,l,mid-1);build(ch[x][1],x,mid+1,r);pushup(x);}inline void Init(){tot=root=0;ch[root][0]=ch[root][1]=pre[root]=size[root]=0;rev[root]=add[root]=0;m[root]=inf;newnode(root,0,inf);newnode(ch[root][1],root,inf);build(key_root,ch[root][1],1,n);pushup(ch[root][1]);pushup(root);}inline void Rotate(int x,int kind){int y=pre[x];pushdown(y);pushdown(x);//pr(root);cout<<"RRRRRR"<<endl;system("pause");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];//cout<<y<<" "<<x<<" "<<ch[x][kind]<<endl;ch[x][kind]=y;pre[y]=x;pushup(y);pushup(x);}inline void Splay(int r,int goal){pushdown(r);//pr(root);system("pause");while(pre[r]!=goal){if(pre[pre[r]]==goal){pushdown(pre[r]);pushdown(r);Rotate(r,ch[pre[r]][0]==r);}else{pushdown(pre[pre[r]]);pushdown(pre[r]);pushdown(r);int y=pre[r];int kind=(ch[pre[y]][0]==y);if(ch[y][kind]==r){Rotate(r,!kind);Rotate(r,kind);}else{Rotate(y,kind);//pr(root);cout<<"SSSSS"<<endl;//cout<<root<<endl<<ch[root][0]<<" "<<ch[root][1]<<" "<<key_root<<" "<<ch[ch[root][1]][0]<<endl;//system("pause");Rotate(r,kind);}}}//system("pause");pushup(r);if(goal==0)root=r;}int Get_key(int r,int k){pushdown(r);int t=size[ch[r][0]]+1;if(t==k)return r;else if(t>k)return Get_key(ch[r][0],k);else return Get_key(ch[r][1],k-t);}inline void Add(int l,int r,int k){//pr(root);//cout<<Get_key(root,l)<<endl;system("pause");Splay(Get_key(root,l),0);//pr(root);Splay(Get_key(root,r+2),root);update_add(key_root,k);pushdown(key_root);pushup(ch[root][1]);pushup(root);}inline void Rev(int l,int r){Splay(Get_key(root,l),0);Splay(Get_key(root,r+2),root);update_rev(key_root);pushdown(key_root);pushup(ch[root][1]);pushup(root);}inline void Delete(int x){Splay(Get_key(root,x),0);Splay(Get_key(root,x+2),root);pre[key_root]=0;key_root=0;pushup(ch[root][1]);pushup(root);}inline void Insert(int x,int c){Splay(Get_key(root,x+1),0);Splay(Get_key(root,x+2),root);newnode(key_root,ch[root][1],c);pushup(ch[root][1]);pushup(root);}inline int Get_min(int l,int r){Splay(Get_key(root,l),0);Splay(Get_key(root,r+2),root);return m[key_root];}inline void Revolve(int l,int r,int t){if(t==0)return;int k=r-t;Splay(Get_key(root,l),0);Splay(Get_key(root,k+2),root);int tmp=key_root;key_root=0;pushup(ch[root][1]);pushup(root);Splay(Get_key(root,r-k+l),0);Splay(Get_key(root,r-k+l+1),root);key_root=tmp;pre[key_root]=ch[root][1];pushup(ch[root][1]);pushup(root);}int main(){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);Init();int T;scanf("%d",&T);char str[10];int l,r,k;while(T--){scanf("%s",str);if(!strcmp(str,"ADD")){scanf("%d%d%d",&l,&r,&k);Add(l,r,k);}else if(!strcmp(str,"REVERSE")){scanf("%d%d",&l,&r);Rev(l,r);}else if(!strcmp(str,"REVOLVE")){scanf("%d%d%d",&l,&r,&k);int b=r-l+1;Revolve(l,r,(k%b+b)%b);//pr(root);}else if(!strcmp(str,"INSERT")){scanf("%d%d",&l,&k);Insert(l,k);//pr(root);}else if(!strcmp(str,"DELETE")){scanf("%d",&l);Delete(l);}else{scanf("%d%d",&l,&r);printf("%d\n",Get_min(l,r));}}}