POJ 1442 Black Box

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

Northeastern Europe 1996

#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 YcdoiT. ( style: Codeblocks )