POJ 1442 -- Black Box(大小堆，优先队列)

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



题解： 用最大堆保存每次取出的最小数，然后用当前最小堆队列的最小值和最大堆的比较，如果大于最大堆需要两个值互换， 应该最大堆的top保存的是第k-1小的数，如果最小堆有比它小的，那么最大堆的top应变成第k小的数。

#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<cstring>
using namespace std;

const int maxn = 30100;

int num[maxn];
priority_queue<int> big;
priority_queue<int,vector<int>,greater<int> > small;

int main(){
    
    int n,m,op;
    
    scanf("%d%d",&n,&m);
    
    for(int i=0;i<n;i++){
        scanf("%d",&num[i]);
    }
    int cnt = 0;
    for(int i=0;i<m;i++){
        scanf("%d",&op);
        
        while(cnt<op){
            small.push(num[cnt]);
        
            if(!big.empty()&&big.top()>small.top()){
                int tmp1 = big.top();
                int tmp2 = small.top();
                big.pop();
                small.pop();
                big.push(tmp2);
                small.push(tmp1);
            }
            cnt++;
        }
        printf("%d\n",small.top());
        big.push(small.top());
        small.pop();                
    }
    
    return 0;
}