poj 3580(Splay)

SuperMemo

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 14372		Accepted: 4489
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:

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

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

代码上参照了cxlove大神的版，原文链接：http://blog.csdn.net/acm_cxlove/article/details/7803646
splay模板题;
参考：OI2007余江伟《如何处理好动态统计问题》
OI2004 杨思雨《伸展树的基本操作和应用》
平衡树之splay讲解 http://www.cnblogs.com/BLADEVIL/p/3464458.html 另外还有维基百科上关于Splay的讲解

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#define MAXN 200005
#define inf 1<<29
#define key_value (ch[ch[root][1]][0])
using namespace std;
int n,q,a[MAXN];
int m[MAXN],rev[MAXN],size[MAXN],add[MAXN],pre[MAXN],ch[MAXN][2],val[MAXN];
int tot1,tot2,root,s[MAXN];
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;
m[r]+=ADD;
val[r]+=ADD;
add[r]+=ADD;
}
void newnode(int &r,int k,int father)
{
if(tot2)
r=s[tot2--];
else
r=++tot1;
ch[r][0]=ch[r][1]=0;
pre[r]=father;
val[r]=m[r]=k;
size[r]=1;
add[r]=0;
}
void push_up(int x)
{
int l=ch[x][0],r=ch[x][1];
size[x]=size[l]+size[r]+1;
m[x]=min(min(m[l],m[r]),val[x]);
}
void push_down(int x)
{
int l=ch[x][0],r=ch[x][1];
if(add[x])
{
update_add(l,add[x]);
update_add(r,add[x]);
add[x]=0;
}
if(rev[x])
{
update_rev(l);
update_rev(r);
rev[x]=0;
}
}
void build(int &r,int L,int R,int father)
{
if(L>R)
return;
int mid=(L+R)/2;
newnode(r,a[mid],father);
build(ch[r][0],L,mid-1,r);
build(ch[r][1],mid+1,R,r);
push_up(r);
}
void Init()
{
tot1=tot2=root=0;
ch[root][0]=ch[root][1]=pre[root]=rev[root]=size[root]=add[root]=0;
m[root]=inf;
newnode(root,inf,0);
newnode(ch[root][1],inf,root);
push_up(root);
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];
ch[x][kind]=y;
pre[y]=x;
push_up(y);
}
void Splay(int r,int goal)
{
push_down(r);
while(pre[r]!=goal)
{
if(pre[pre[r]]==goal)
rotate(r,ch[pre[r]][0]==r);
else
{
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);
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;
}
void reversal(int l,int r)
{
int x=get_kth(root,l);
Splay(x,0);
int y=get_kth(root,r+2);
Splay(y,root);
update_rev(key_value);
}
void insert(int pos,int k)
{
int x=get_kth(root,pos);
Splay(x,0);
int y=get_min(ch[root][1]);
Splay(y,root);
newnode(key_value,k,ch[root][1]);
push_up(ch[root][1]);
push_up(root);
}
void erase(int r)
{
if(!r) return;
s[++tot2]=r;
erase(ch[r][0]);
erase(ch[r][1]);
}
void ADD(int l,int r,int k)
{
int x=get_kth(root,l);
Splay(x,0);
int y=get_kth(root,r+2);
Splay(y,root);
update_add(key_value,k);
push_up(ch[root][1]);
push_up(root);
}
int ans(int l,int r)
{
int x=get_kth(root,l);
Splay(x,0);
int y=get_kth(root,r+2);
Splay(y,root);
return m[key_value];
}
void revolve(int a,int b,int t)
{
if(!t)
return;
int c=b-t;
int x=get_kth(root,a);
Splay(x,0);
x=get_kth(root,c+2);
Splay(x,root);
int tmp=key_value;
key_value=0;
push_up(ch[root][1]);
push_up(root);
Splay(get_kth(root,b-c+a),0);
Splay(get_kth(root,b-c+a+1),root);
key_value=tmp;
pre[key_value]=ch[root][1];
push_up(ch[root][1]);
push_up(root);
}
void Delete(int l)
{
int x=get_kth(root,l);
Splay(x,0);
int y=get_kth(root,l+2);
Splay(y,root);
erase(key_value);
pre[key_value]=0;
key_value=0;
push_up(ch[root][1]);
push_up(root);
}
int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
scanf("%d",&q);
Init();
char str[10];
int l,r,k;
while(q--){
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);
reversal(l,r);
}
else if(!strcmp(str,"REVOLVE")){
scanf("%d%d%d",&l,&r,&k);
revolve(l,r,(k%(r-l+1)+(r-l+1))%(r-l+1));
}
else if(!strcmp(str,"INSERT")){
scanf("%d%d",&l,&k);
insert(l+1,k);
}
else if(!strcmp(str,"DELETE")){
scanf("%d",&l);
Delete(l);
}
else{
scanf("%d%d",&l,&r);
printf("%d\n",ans(l,r));
}
}
}
return 0;
}