POJ 1442 动态第k大

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 6365		Accepted: 2580

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

treap入门学习，直接上代码：

/* ***********************************************
Author :xianxingwuguan
Created Time :2014/3/1 14:54:52
File Name :5.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
const int maxn=100100;
struct Treap{
	int root,treapnode,key[maxn*5],priority[maxn*5];
	int ch[maxn*5][2],cnt[maxn*5],size[maxn];
	Treap(){
		root=0;
		treapnode=1;
		priority[0]=INF;
		size[0]=0;
	}
	void up(int x){
		size[x]=size[ch[x][0]]+cnt[x]+size[ch[x][1]];
	}
	void rotate(int &x,int t){
		int y=ch[x][t];
		ch[x][t]=ch[y][t^1];
		ch[y][t^1]=x;
		up(x);up(y);x=y;
	}
	void insert(int &x,int k){
		if(x){
			if(key[x]==k)cnt[x]++;
			else {
				int t=key[x]<k;
				insert(ch[x][t],k);
				if(priority[ch[x][t]]<priority[x])
					rotate(x,t);
			}
		}
		else {
			x=treapnode++;
			key[x]=k;
			cnt[x]=1;
			priority[x]=rand();
			ch[x][0]=ch[x][1]=0;
		}
		up(x);
	}
	int getkth(int &x,int k){
		if(k<=size[ch[x][0]])return getkth(ch[x][0],k);
		k-=size[ch[x][0]]+cnt[x];
		if(k<=0)return key[x];
		return getkth(ch[x][1],k);
	}
}T;
int M,N,II,cc;
int arr[33000];
int u[33000];

int main()
{
    while(~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(T.root,arr[i]);
           while(i==u[cc]){
            cc++;II++;
            printf("%d\n",T.getkth(T.root,II));
          }
     }
  }
    return 0;
}