poj1442 Black Box treap

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Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6815 Accepted: 2753

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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

Source

Northeastern Europe 1996

题意：给定一组数据按顺序插入，每次在u个元素的时候，找到第i小的数，i= 1，2，3，4...，

思路：treap求第k小，详见代码：

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int MAXN=30000+100;int n,m;int a[MAXN];struct node{    node *ch[2];    int r,v,s;    node(int v):v(v)    {        s=1;        r=rand();        ch[0]=ch[1]=NULL;    }    bool operator <(const node &rhs) const    {        return r<rhs.r;    }    int cmp(int x)    {        if(x==v)            return -1;        return x<v ? 0:1;    }    void maintain()    {        s=1;        if(ch[0]!=NULL) s+=ch[0]->s;        if(ch[1]!=NULL) s+=ch[1]->s;    }};struct Treap{    inline void rotate(node * &o,int d)    {        node *k=o->ch[d^1];        o->ch[d^1]=k->ch[d];        k->ch[d]=o;        o->maintain(); k->maintain();        o=k;    }    inline void insert(node * &o,int x)    {        if(o==NULL) o=new node(x);        else        {            int d=(x<o->v?0:1);            insert(o->ch[d],x);            if(o->ch[d]->r>o->r)                rotate(o,d^1);        }        o->maintain();    }    inline void remove(node * &o,int x)    {        int d=o->cmp(x);        if(d==-1)        {            node *u=o;            if(o->ch[0]!=NULL && o->ch[1]!=NULL)            {                int d2=(o->ch[0]->r>o->ch[1]->r? 1:0);                rotate(o,d2);remove(o->ch[d2],x);            }            else            {                if(o->ch[0]==NULL)                    o=o->ch[1];                else                    o=o->ch[0];                delete u;            }        }        else            remove(o->ch[d],x);        if(o!=NULL)            o->maintain();    }    inline int kth(node * o,int k)    {        if(o==NULL || k<=0 || k>o->s) return 0;        int s=(o->ch[0]==NULL ? 0: o->ch[0]->s);        if(k==s+1) return o->v;        else if(k<=s) return kth(o->ch[0],k);        else return kth(o->ch[1],k-s-1);    }}treap;int main(){    //freopen("text.txt","r",stdin);    scanf("%d%d",&n,&m);    node *rt=NULL;    for(int i=1;i<=n;i++)        scanf("%d",&a[i]);    int st=1;    for(int i=1;i<=m;i++)    {        int x;        scanf("%d",&x);        while(st<=x)        {            treap.insert(rt,a[st]);            st++;        }        printf("%d\n",treap.kth(rt,i));    }    return 0;}