HDU4453(伸展树模板题)
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LooploopTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1859 Accepted Submission(s): 607
Problem Description
XXX gets a new toy named Looploop. The toy has N elements arranged in a loop, an arrow pointing to one of the elements, and two preset parameters k1 and k2. Every element has a number on it.
The figure above shows a Looploop of 6 elments. Let's assuming the preset parameter k1 is 3, and k2 is 4. XXX can do six operations with the toy. 1: add x Starting from the arrow pointed element, add x to the number on the clockwise first k2 elements. 
2: reverse Starting from the arrow pointed element, reverse the first k1 clockwise elements. 
3: insert x Insert a new element with number x to the right (along clockwise) of the arrow pointed element. 
4: delete Delete the element the arrow pointed and then move the arrow to the right element. 
5: move x x can only be 1 or 2. If x = 1 , move the arrow to the left(along the counterclockwise) element, if x = 2 move the arrow to the right element. 
6: query Output the number on the arrow pointed element in one line. 
XXX wants to give answers to every query in a serial of operations.
Input
There are multiple test cases.
For each test case the first line contains N,M,k1,k2(2≤k1 Second line contains N integers ai(-10 4≤ai≤10 4) representing the N numbers on the elements in Looploop along clockwise direction. The arrow points to first element in input at the beginning. Then m lines follow, each line contains one of the six operations described above. It is guaranteed that the "x" in the "add","insert" and "move" operations is always integer and its absolute value ≤104. The number of elements will never be less than N during the operations. The input ends with a line of 0 0 0 0.
Output
For each test case, output case number in the first line(formatted as the sample output). Then for each query in the case, output the number on the arrow pointed element in a single line.
Sample Input
5 1 2 4 3 4 5 6 7 query 5 13 2 4 1 2 3 4 5 move 2 query insert 8 reverse query add 2 query move 1 query move 1 query delete query 0 0 0 0
Sample Output
Case #1: 3 Case #2: 2 8 10 1 5 1
Source
2012 Asia Hangzhou Regional Contest
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#include
#include
#include
#include
#include
using namespace std;
const int maxn=2*100010;
int root;
int N;
int ch[maxn][2];
int pre[maxn];
int a[maxn];
int size[maxn];
int key[maxn];
int add[maxn];
int rev[maxn];
int s[maxn];
int tot2;
int tot1;
#define Key_value ch[ch[root][1]][0]
void NewNode(int &r,int f,int val)
{
if(tot2)r=s[tot2--];
else r+=++tot1;
ch[r][0]=ch[r][1]=0;
pre[r]=f;
size[r]=1;
key[r]=val;
rev[r]=0;
add[r]=0;
}
void pushup(int r)
{
size[r]=size[ch[r][0]]+size[ch[r][1]]+1;
}
void build(int &x,int l,int r,int f)
{
if(l>r)return ;
int mid=(l+r)>>1;
NewNode(x,f,a[mid]);
build(ch[x][0],l,mid-1,x);
build(ch[x][1],mid+1,r,x);
pushup(x);
}
void init()
{
root=tot1=tot2=0;
ch[root][0]=ch[root][1]=size[root]=key[root]=add[root]=pre[root]=0;
rev[root]=0;
NewNode(root,0,-1);
NewNode(ch[root][1],root,-1);
build(Key_value,1,N,ch[root][1]);
pushup(ch[root][1]);
pushup(root);
}
void update_add(int r,int x)
{
if(!r)return ;
add[r]+=x;
key[r]+=x;
}
void update_rev(int r)
{
if(!r)return ;
swap(ch[r][1],ch[r][0]);
rev[r]^=1;
}
void pushdown(int r) //从r号结点开始向下更新
{
if(add[r]) //如果add数组可以更新
{
update_add(ch[r][0],add[r]);
update_add(ch[r][1],add[r]);
add[r]=0;
}
if(rev[r]) //如果rev数组可以更新
{
update_rev(ch[r][0]);
update_rev(ch[r][1]);
rev[r]=0;
}
}
void Rotate(int x,int kind)
{
int y=pre[x];
pushdown(y);
pushdown(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;
pushup(y);
}
void Splay(int r,int goal)
{
pushdown(r);
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);
Rotate(r,kind);
}
}
}
pushup(r);
if(goal==0)root=r;
}
int get_kth(int r,int k)
{
pushdown(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);
}
void change(int &poss,int k) //让数组中[pos,k]的数放入可被更改
{ //的一棵树中
int pos=get_kth(root,poss); // (根节点序号为Key_value)
Splay(pos,0);
int tmp=size[ch[root][0]]+1;
Splay(get_kth(root,1),0); //就位
Splay(get_kth(root,tmp),root);
tmp=Key_value;
Key_value=0;
pushup(ch[root][1]);
pushup(root);
Splay(get_kth(root,size[root]-1),0);
Key_value=tmp;
pre[Key_value]=ch[root][1];
pushup(ch[root][1]);
pushup(root);
Splay(get_kth(root,1),0);
Splay(get_kth(root,k+2),root);
}
void Add(int &poss,int k2,int x) //[poss,poss+k2]区间所有值加x
{
change(poss,k2);
update_add(Key_value,x);
pushup(ch[root][1]);
pushup(root);
poss=2;
}
void Reverse(int &poss,int k1) //翻转[poss,poss+k1]的值
{
change(poss,k1);
update_rev(Key_value);
pushup(ch[root][1]);
pushup(root);
poss=2;
}
//在序号为pos处插入x
void Insert(int &poss,int x)
{
//将pos放到0号点下方
int pos=get_kth(root,poss);
Splay(pos,0);
//
Splay(get_kth(root,size[ch[root][0]]+2),root);
NewNode(Key_value,ch[root][1],x);
pushup(ch[root][1]);
pushup(root);
}
void erase(int r)
{
if(r)
{
s[++tot2]=r;
erase(ch[r][0]);
erase(ch[r][1]);
}
}
//删除操作
void Delete(int &poss)
{
int pos=get_kth(root,poss);
Splay(pos,0);
int tmp=size[ch[root][0]]+1;
//就位
Splay(get_kth(root,tmp-1),0);
Splay(get_kth(root,tmp+1),root);
//清除
erase(Key_value);
pre[Key_value]=0;
Key_value=0;
//更新
pushup(ch[root][1]);
//更新
pushup(root);
//更新指针
if(tmp==size[root])poss=2;
else poss=tmp;
}
//光标移动
void Move(int &poss,int x)
{
int pos=get_kth(root,poss);
if(x==1)
{
Splay(pos,0);
int tmp=size[ch[root][0]];
if(size[ch[root][0]]==1)tmp=size[root]-1;
poss=tmp;
//pos=get_kth(root,tmp);
}
else
{
Splay(pos,0);
int tmp=size[ch[root][0]]+2;
if(tmp==size[root])tmp=2;
poss=tmp;
//pos=get_kth(root,tmp);
}
}
//询问
int Query(int &poss)
{
int pos=get_kth(root,poss);
Splay(pos,0);
return key[root];
}
int main()
{
int n,m,k1,k2;
int t=1;
while(scanf("%d%d%d%d",&n,&m,&k1,&k2)!=EOF)
{
N=n;
if(n==0&&m==0&&k1==0&&k2==0)
break;
printf("Case #%d:\n",t++);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int poss=2;
int sum=n;
init();
for(int i=1;i<=m;i++)
{
char a[50];
scanf("%s",a);
if(a[0]=='q')
{
int res=Query(poss);
printf("%d\n",res);
}
else if(a[0]=='r')
{
Reverse(poss,k1);
}
else if(a[0]=='i')
{
int x;
sum+=1;
scanf("%d",&x);
Insert(poss,x);
}
else if(a[0]=='d')
{
sum-=1;
Delete(poss);
}
else if(a[0]=='m')
{
int x;
scanf("%d",&x);
if(x==1)
{
if(poss==2)poss=sum+1;
else poss-=1;
}
if(x==2)
{
if(poss==sum+1)poss=2;
else poss+=1;
}
}
else if(a[0]=='a')
{
int x;
scanf("%d",&x);
Add(poss,k2,x);
}
}
}
return 0;
}