第k大数维护，我推荐Treap。。谁用谁知道。。。。

                                                          Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5999		Accepted: 2417

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

Source

 

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

using namespace std;

const int maxNode=444444;

const int INF=0x3f3f3f3f;

int M,N,II,cc;

int arr[33000];

int u[33000];

struct Treap

{

    int root,treapCnt,key[maxNode],priority[maxNode],

    childs[maxNode][2],cnt[maxNode],ssize[maxNode];

    Treap()

    {

        root=0;

        treapCnt=1;

        priority[0]=INF;

        ssize[0]=0;

    }

    void update(int x)

    {

        ssize[x]=ssize[childs[x][0]]+cnt[x]+ssize[childs[x][1]];

    }

    void rotate(int &x,int t)

    {

        int y=childs[x][t];

        childs[x][t]=childs[y][1-t];

        childs[y][1-t]=x;

        update(x);

        update(y);

        x=y;

    }

    void _insert(int &x,int k)

    {

        if(x)

        {

            if(key[x]==k)

            {

                cnt[x]++;

            }

            else

            {

                int t=key[x]<k;

                _insert(childs[x][t],k);

                if(priority[childs[x][t]]<priority[x])

                {

                    rotate(x,t);

                }

            }

        }

        else

        {

            x=treapCnt++;

            key[x]=k;

            cnt[x]=1;

            priority[x]=rand();

            childs[x][0]=childs[x][1]=0;

        }

        update(x);

    }

    int _getKth(int &x,int k)

    {

        if(k<=ssize[childs[x][0]])

        {

            return _getKth(childs[x][0],k);

        }

        k-=ssize[childs[x][0]]+cnt[x];

        if(k<=0)

        {

            return key[x];

        }

        return _getKth(childs[x][1],k);

    }

    void Insert(int k)

    {

        _insert(root,k);

    }

    int GetKth(int k)

    {

        return _getKth(root,k);

    }

}T;

int main()

{

    scanf("%d%d",&M,&N);

    II=0;cc=1;

    for(int i=1;i<=M;i++)

    {

        scanf("%d",arr+i);

    }

    for(int i=1;i<=N;i++)

    {

        scanf("%d",u+i);

    }

    for(int i=1;i<=M;i++)

    {

        T.Insert(arr[i]);

        while(i==u[cc])

        {

            cc++;II++;

            printf("%d\n",T.GetKth(II));

        }

    }

    return 0;

}

* This source code was highlighted by . ( style: Codeblocks )