Treap模板——POJ1442——Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12151		Accepted: 4979

Description

Our Black Box represents a primitive database. It can save an integer array and has a special i variable. At the initial moment Black Box is empty and i equals 0. This Black Box processes a sequence of commands (transactions). There are two types of transactions: 

ADD (x): put element x into Black Box; 
GET: increase i by 1 and give an i-minimum out of all integers containing in the Black Box. Keep in mind that i-minimum is a number located at i-th place after Black Box elements sorting by non- descending. 

Let us examine a possible sequence of 11 transactions: 

Example 1 

N Transaction i Black Box contents after transaction Answer 

      (elements are arranged by non-descending)   

1 ADD(3)      0 3   

2 GET         1 3                                    3 

3 ADD(1)      1 1, 3   

4 GET         2 1, 3                                 3 

5 ADD(-4)     2 -4, 1, 3   

6 ADD(2)      2 -4, 1, 2, 3   

7 ADD(8)      2 -4, 1, 2, 3, 8   

8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8   

9 GET         3 -1000, -4, 1, 2, 3, 8                1 

10 GET        4 -1000, -4, 1, 2, 3, 8                2 

11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8   

It is required to work out an efficient algorithm which treats a given sequence of transactions. The maximum number of ADD and GET transactions: 30000 of each type. 


Let us describe the sequence of transactions by two integer arrays: 


1. A(1), A(2), ..., A(M): a sequence of elements which are being included into Black Box. A values are integers not exceeding 2 000 000 000 by their absolute value, M <= 30000. For the Example we have A=(3, 1, -4, 2, 8, -1000, 2). 

2. u(1), u(2), ..., u(N): a sequence setting a number of elements which are being included into Black Box at the moment of first, second, ... and N-transaction GET. For the Example we have u=(1, 2, 6, 6). 

The Black Box algorithm supposes that natural number sequence u(1), u(2), ..., u(N) is sorted in non-descending order, N <= M and for each p (1 <= p <= N) an inequality p <= u(p) <= M is valid. It follows from the fact that for the p-element of our u sequence we perform a GET transaction giving p-minimum number from our A(1), A(2), ..., A(u(p)) sequence. 

Input

Input contains (in given order): M, N, A(1), A(2), ..., A(M), u(1), u(2), ..., u(N). All numbers are divided by spaces and (or) carriage return characters.

Output

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Sample Input

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

Sample Output

3
3
1
2

给m个数和n的查询，对于第i个查询，找到前i个数里第u[i]大的那个

//by hzwer
#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int MAXN=100010;
struct data{
    //左二子，右儿子，权值，整棵子树大小，优先级，相同权值的节点个数
    int l,r,v,size,rnd,w;
}tr[MAXN];
int n,size,root,ans;
void update(int k)//更新结点信息
{
    tr[k].size=tr[tr[k].l].size+tr[tr[k].r].size+tr[k].w;
}
//右旋
void rturn(int &k)
{
    int t=tr[k].l;tr[k].l=tr[t].r;tr[t].r=k;
    tr[t].size=tr[k].size;update(k);k=t;
}
//左旋
void lturn(int &k)
{
    int t=tr[k].r;tr[k].r=tr[t].l;tr[t].l=k;
    tr[t].size=tr[k].size;update(k);k=t;
}
//插入权值为x的节点
void insert(int &k,int x)
{
    if(k==0)
    {
        size++;k=size;
        tr[k].size=tr[k].w=1;tr[k].v=x;tr[k].rnd=rand();
        return;
    }
    tr[k].size++;
    if(tr[k].v==x)tr[k].w++;
    else if(x>tr[k].v)
    {
        insert(tr[k].r,x);
        if(tr[tr[k].r].rnd<tr[k].rnd)lturn(k);
    }
    else 
    {
        insert(tr[k].l,x);
        if(tr[tr[k].l].rnd<tr[k].rnd)rturn(k);
    } 
}
//删除权值为x的节点，如果有重复的，只删除一个
void del(int &k,int x)
{
    if(k==0)return; 
    if(tr[k].v==x)
    {
        if(tr[k].w>1)
        {
            tr[k].w--;tr[k].size--;return;
        }
        if(tr[k].l*tr[k].r==0)k=tr[k].l+tr[k].r;
        else if(tr[tr[k].l].rnd<tr[tr[k].r].rnd)
            rturn(k),del(k,x);
        else lturn(k),del(k,x);
    }
    else if(x>tr[k].v)
        tr[k].size--,del(tr[k].r,x);
    else tr[k].size--,del(tr[k].l,x);
}
//查询x的排名，如果有多个，输出最小的一个
int query_rank(int k,int x)
{
    if(k==0)return 0;
    if(tr[k].v==x)return tr[tr[k].l].size+1;
    else if(x>tr[k].v)
        return tr[tr[k].l].size+tr[k].w+query_rank(tr[k].r,x);
    else return query_rank(tr[k].l,x);
}
//查询排名为x的数
int query_num(int k,int x)
{
    if(k==0)return 0;
    if(x<=tr[tr[k].l].size)
        return query_num(tr[k].l,x);
    else if(x>tr[tr[k].l].size+tr[k].w)
        return query_num(tr[k].r,x-tr[tr[k].l].size-tr[k].w);
    else return tr[k].v;
}
//求x的前驱
void query_pro(int k,int x)
{
    if(k==0)return;
    if(tr[k].v<x)
    {
        ans=k;query_pro(tr[k].r,x);
    }
    else query_pro(tr[k].l,x);
}
//求x的后继
void query_sub(int k,int x)
{
    if(k==0)return;
    if(tr[k].v>x)
    {
        ans=k;query_sub(tr[k].l,x);
    }
    else query_sub(tr[k].r,x);
}
// int main()
// {
//     scanf("%d",&n);
//     int opt,x;
//     for(int i=1;i<=n;i++)
//     {
//         scanf("%d%d",&opt,&x);
//         switch(opt)
//         {
//         case 1:insert(root,x);break;
//         case 2:del(root,x);break;
//         case 3:printf("%d\n",query_rank(root,x));break;
//         case 4:printf("%d\n",query_num(root,x));break;
//         case 5:ans=0;query_pro(root,x);printf("%d\n",tr[ans].v);break;
//         case 6:ans=0;query_sub(root,x);printf("%d\n",tr[ans].v);break;
//         }
//     }
//     return 0;
// }

int a[MAXN],u[MAXN];

int main()
{
    int n,m;
    scanf("%d%d",&m,&n);
    for(int i=1;i<=m;i++){
        scanf("%d",a+i);
    }
    for(int i=1;i<=n;i++){
        scanf("%d",u+i);
    }
    int i=1,j=1;
    while(j<=n){
        while(i<=u[j]){
            insert(root,a[i++]);
        }
        printf("%d\n",query_num(root,j++));
    }
}