POJ 1442-Black Box（动态区间第K小-优先队列）

秒速五厘米 2022-07-11 13:27 136阅读 0赞

Black Box

<table> 
 <tbody> 
  <tr> 
   <td><strong>Time Limit:</strong>&nbsp;1000MS</td> 
   <td>&nbsp;</td> 
   <td><strong>Memory Limit:</strong>&nbsp;10000K</td> 
  </tr> 
  <tr> 
   <td><strong>Total Submissions:</strong>&nbsp;12404</td> 
   <td>&nbsp;</td> 
   <td><strong>Accepted:</strong>&nbsp;5088</td> 
  </tr> 
 </tbody> 
</table>

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][]

## 题目意思： ##

初始情况下，序列中没有任何元素（题中序列第一个元素K表示要GET的第K小元素，不算是序列中的元素）。

ADD(X)是将元素X加入序列中；GET是将K加一后，从序列中取出第K小元素。

给出M个A数组中的元素，表示ADD中的X；N个U数组中的元素，表示当前序列中的元素总个数。

要求出每次GET从序列中取出第K小元素。

## 解题思路： ##

q是最大堆的优先队列，存储序列的最小元素到序列的第K小元素；

p是最小堆的优先队列，存储序列的第K+1小元素到序列的最大元素。

**压入元素：**

遍历U数组，确定当前A数组中要压入队列的元素个数，分情况讨论：

①q中元素不足k个

注意这里不是直接将元素压入q，而是先将元素压入p，再从q里取出最小值压入q。

②q中元素大于等于k个

1’：当前要压入的元素比q的最大值小，q的最大值压入p后出队，当前元素压入q；

2’：当前要压入的元素比q的最大值大（或等于），直接将当前元素压入q。

**输出元素：**

如果q数组中元素个数小于k，依次将p的队首元素压入q，直到q数组中元素个数等于k。

否则直接输出q的队首元素。

#include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    #include <algorithm>
    const int MAXN=30030;
    int a[MAXN],u[MAXN];
    using namespace std;
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int m,n;
        cin>>m>>n;
        for(int i=0; i<m; ++i)
            cin>>a[i];
        for(int i=0; i<n; ++i)
            cin>>u[i];
        priority_queue<int,vector<int>,greater<int> >p;//最小值优先
        priority_queue<int>q;//最大值优先
        int k=0,ac=0;//第k小、a数组中的指针
        for(int i=0; i<n; ++i)//遍历u数组
        {
            ++k;
            int ps=p.size(),qs=q.size();
            for(int j=0; j<u[i]-ps-qs; ++j)//向队列中压入的a数组的元素个数
            {
                if(q.size()<k)//q中元素不足k个
                {
                    //先将元素压入p，再从q里取出最小值压入q
                    p.push(a[ac]);
                    q.push(p.top());
                    p.pop();
                    ++ac;
                }
                else//q中元素大于k个
                {
                    if(q.top()>a[ac])//当前要压入的元素比q的最大值小
                    {
                        //q的最大值压入p后出队，当前元素压入q
                        p.push(q.top());
                        q.pop();
                        q.push(a[ac]);
                        ++ac;
                    }
                    else//当前要压入的元素比q的最大值大
                    {
                        //直接将当前元素压入q
                        p.push(a[ac]);
                        ++ac;
                    }
                }
            }
            if(q.size()<k)//如果q数组中元素个数小于k
            {
                //依次将p的队首元素压入q，直到q数组中元素个数等于k
                int temp=k-q.size();//需要压入q的元素个数
                for(int j=0; j<temp; ++j)
                {
                    q.push(p.top());
                    p.pop();
                }
            }
            cout<<q.top()<<endl;
        }
        return 0;
    }
    /*
    7 4
    3 1 -4 2 8 -1000 2
    1 2 6 6
    */

[Northeastern Europe 1996]: http://poj.org/searchproblem?field=source&key=Northeastern+Europe+1996