[POJ 1442] Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7770		Accepted: 3178

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

 

#include <iostream>
#include <stdio.h>
#include <time.h>
using namespace std;
#define N 30100
#define INF 0x7fffffff

int val[N];
struct Treap
{
    int ch[N][2],fix[N],size[N];
    int top,root;
    
    inline void PushUp(int rt)
    {
        size[rt]=size[ch[rt][0]]+size[ch[rt][1]]+1;
    }
    void Newnode(int &rt,int v)
    {
        rt=++top;
        val[rt]=v;
        size[rt]=1;
        fix[rt]=rand();
        ch[rt][0]=ch[rt][1]=0;
    }
    void Init()
    {
        top=root=0;
        ch[0][0]=ch[0][1]=0;
        size[0]=val[0]=0;
        memset(size,0,sizeof(size));
    }
    void Rotate(int &x,int kind)
    {
        int y=ch[x][kind^1];
        ch[x][kind^1]=ch[y][kind];
        ch[y][kind]=x;
        PushUp(x);
        PushUp(y);
        x=y;
    }
    void Insert(int &rt,int v)
    {
        if(!rt)
            Newnode(rt,v);
        else
        {
            int kind=(v>=val[rt]);
            size[rt]++;
            Insert(ch[rt][kind],v);
            if(fix[ch[rt][kind]]<fix[rt])
                Rotate(rt,kind^1);
        }
    }
    int Find(int rt,int k)
    {
        int cnt=size[ch[rt][0]];
        if(k==cnt+1) return val[rt];
        else if(k<=cnt) return Find(ch[rt][0],k);
        else return Find(ch[rt][1],k-cnt-1);
    }
}t;
int main()
{
    int n,i,m;
    srand((int)time(0));
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        t.Init();
        for(i=1;i<=n;i++)
        {
            scanf("%d",&val[i]);
        }
        int last=1,x;
        for(i=1;i<=m;i++)
        {
            scanf("%d",&x);
            while(last<=x)
            {
                t.Insert(t.root,val[last]);
                last++;
            }
            printf("%d
",t.Find(t.root,i));
        }
    }
    return 0;
}