N的范围很大,但Q的范围比较小.可以把TOP,QUERY操作用到的点分离出来,没用到的段缩成点

对于TOP 把x转到根,删除后加到开头位置

对于QUERY 旋转到根直接输出

对于RANK,递归

Queue-jumpers

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Ponyo and Garfield are waiting outside the box-office for their favorite movie. Because queuing is so boring, that they want to play a game to kill the time. The game is called “Queue-jumpers”. Suppose that there are N people numbered from 1 to N stand in a
line initially. Each time you should simulate one of the following operations:

1.  Top x :Take person x to the front of the queue

2.  Query x: calculate the current position of person x

3.  Rank x: calculate the current person at position x

Where x is in [1, N].

Ponyo is so clever that she plays the game very well while Garfield has no idea. Garfield is now turning to you for help.

Input

In the first line there is an integer T, indicates the number of test cases.(T<=50)

In each case, the first line contains two integers N(1<=N<=10^8), Q(1<=Q<=10^5). Then there are Q lines, each line contain an operation as said above.

Output

For each test case, output “Case d:“ at first line where d is the case number counted from one, then for each “Query x” operation ,output the current position of person x at a line, for each “Rank x” operation, output the current person at position x at a line.

Sample Input

3
9 5
Top 1
Rank 3
Top 7
Rank 6
Rank 8
6 2
Top 4
Top 5
7 4
Top 5
Top 2
Query 1
Rank 6

Sample Output

Case 1:
3
5
8
Case 2:
Case 3:
3
6

Author

wzc1989

Source

2010 ACM-ICPC
Multi-University Training Contest（1）——Host by FZU

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=200200;

char op[maxn][10];
int number[maxn];
int n;
int s[maxn],e[maxn];
int p[maxn];

/***************SPLAY*********************/

int ch[maxn][2],num[maxn],sz[maxn],pre[maxn];
int tot1,root;

void NewNode(int& x,int father,int k)
{
    x=k;
    sz[x]=num[x]=e[x]-s[x]+1;
    ch[x][1]=ch[x][0]=0;
    pre[x]=father;
}

void Push_Up(int x)
{
    sz[x]=sz[ch[x][0]]+sz[ch[x][1]]+num[x];
}

void Build(int& x,int l,int r,int fa)
{
    if(l>r) return ;
    int mid=(l+r)/2;
    NewNode(x,fa,mid);
    Build(ch[x][0],l,mid-1,x);
    Build(ch[x][1],mid+1,r,x);
    Push_Up(x);
}

void Init()
{
    root=0;
    ch[root][0]=ch[root][1]=pre[root]=sz[root]=num[root]=0;
    Build(root,1,n,0);
    Push_Up(root);
}

void Rotate(int x,int kind)
{
    int y=pre[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]; pre[y]=x;
    ch[x][kind]=y;
    Push_Up(y);
}

void Splay(int r,int goal)
{
    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);
            else Rotate(y,kind);
            Rotate(r,kind);
        }
    }
    Push_Up(r);
    if(goal==0) root=r;
}

int Get_Min(int r)
{
    while(ch[r][0]) r=ch[r][0];
    return r;
}

void Remove_Root()
{
    if(ch[root][1]==0||ch[root][0]==0)
    {
        root=ch[root][1]+ch[root][0];
        pre[root]=0;
        return ;
    }
    int k=Get_Min(ch[root][1]);
    Splay(k,root);
    ch[ch[root][1]][0]=ch[root][0];
    root=ch[root][1];
    pre[ch[root][0]]=root;
    pre[root]=0;
    Push_Up(root);
}

int Bin(int x)
{
    int low=1,high=n,mid;
    while(low<=high)
    {
        mid=(low+high)/2;
        if(s[mid]<=x&&x<=e[mid]) return mid;
        if(s[mid]>x) high=mid-1;
        else low=mid+1;
    }
    return -1;
}

/************doit*********/

void TOP(int x)
{
    int y=Bin(x);
    Splay(y,0);
    Remove_Root();
    Splay(Get_Min(root),0);
    ch[y][0]=0; ch[y][1]=root;
    pre[root]=y;root=y;
    pre[root]=0;
    Push_Up(root);
}

int QUERY(int x)
{
    int y=Bin(x);
    Splay(y,0);
    return sz[ch[root][0]]+1;
}

int RANK(int r,int x)
{
    int t=sz[ch[r][0]];
    if(x<=t) return RANK(ch[r][0],x);
    else if(x<=t+num[r])
    {
        return s[r]+x-t-1;
    }
    else return RANK(ch[r][1],x-t-num[r]);
}

/*********DEBUG***********/
void showit(int x)
{
    if(x)
    {
        showit(ch[x][0]);
        printf("结点: %2d 左儿子: %2d 右儿子: %2d 父结点: %2d size: %2d num: %2d \n",
               x,ch[x][0],ch[x][1],pre[x],sz[x],num[x]);
        showit(ch[x][1]);
    }
}

void debug()
{
    cout<<"root : "<<root<<endl;
    showit(root);
}

int main()
{
    int T_T,cas=1;
    scanf("%d",&T_T);
while(T_T--)
{
    printf("Case %d:\n",cas++);
    int t=0;
    int nn,mm;
    scanf("%d%d",&nn,&mm);
    for(int i=0;i<mm;i++)
    {
        scanf("%s%d",op[i],number+i);
        if(op[i][0]==‘T‘||op[i][0]==‘Q‘)
        {
            p[t++]=number[i];
        }
    }

    n=0;
    p[t++]=1;p[t++]=nn;
    sort(p,p+t);
    t=unique(p,p+t)-p;

    for(int i=0;i<t;i++)
    {
        if(i!=0&&p[i]-1!=p[i-1])
        {
            n++; s[n]=p[i-1]+1; e[n]=p[i]-1;
        }
        n++; s[n]=e[n]=p[i];
    }

    Init();
    for(int i=0;i<mm;i++)
    {
        if(op[i][0]==‘T‘) TOP(number[i]);
        else if(op[i][0]==‘Q‘)
            printf("%d\n",QUERY(number[i]));
        else printf("%d\n",RANK(root,number[i]));
    }
}
    return 0;
}

HDOJ 3436 Queue-jumpers,布布扣,bubuko.com