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POJ 1442-Black Box(优先队列)
2015-07-20 17:45:48 来源: 作者: 【 】 浏览:3
Tags:POJ 1442-Black Box 优先 队列
Black Box
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 7436 Accepted: 3050

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
题意 :给出n,m,然后接下一行输入n个数,最后一行输入m个数,要求对于最后一行的第i个数,输出在原数列中前x个数中第i小的数,。好绕口,然后一开始用sort是各种TLE,好多大神用平衡树过的,差点没吓哭我,然后,。。撸了两个堆,一个最大堆,一个最小堆,注意维护好这两个堆,其中最大堆放前k个最小的数,最小堆放剩余的数,那么最大堆的堆顶就是第k个最小的数。
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