Treap [poj 1442] Black Box

Black Box

Time limit:1000 ms		Memory limit:10000 K
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 encoded 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 encoded 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 43 1 -4 2 8 -1000 21 2 6 6

Sample output

3312

Source

Northeastern Europe 1996.

#include <iostream>#include <stdio.h>#include <time.h>using namespace std;#define N 30100#define INF 0x7fffffffint 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\n",t.Find(t.root,i));        }    }    return 0;}

 

Treap [poj 1442] Black Box