POJ 1442 Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 5165		Accepted: 2091

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

今天刚学习了优先队列，终于解决这个知识点了

#include <iostream>
#include <stdio.h>
#include <queue>
#include <vector>
using namespace std;
int a[31000],b[31000];
int main()
{
    int i,j,n,m,s,t;
    scanf("%d %d",&n,&m);
    for(i=0;i<=n-1;i++)
    {
        scanf("%d",&a[i]);
    }
    for(j=1;j<=m;j++)
    {
        scanf("%d",&b[j]);
    }
    b[0]=0;
    priority_queue< int,vector<int>,greater<int> > q1;
    priority_queue< int,vector<int>, less<int> > q2;
    for(i=1;i<=m;i++)
    {
        for(j=b[i-1];j<=b[i]-1;j++)
        {
            q1.push(a[j]);
            if(!q2.empty()&&q1.top()<q2.top())
            {
                q1.push(q2.top());
                q2.pop();
                q2.push(q1.top());
                q1.pop();
            }
        }
        q2.push(q1.top());
        q1.pop();
        printf("%d\n",q2.top());
    }
    return 0;
}