POJ 1442-Black Box（動態區間第K小-優先佇列）

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 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.

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

3
3
1
2

Northeastern Europe 1996

初始情況下，序列中沒有任何元素（題中序列第一個元素K表示要GET的第K小元素，不算是序列中的元素）。

ADD(X)是將元素X加入序列中；GET是將K加一後，從序列中取出第K小元素。

給出M個A陣列中的元素，表示ADD中的X；N個U陣列中的元素，表示當前序列中的元素總個數。

要求出每次GET從序列中取出第K小元素。

q是最大堆的優先佇列，儲存序列的最小元素到序列的第K小元素；

p是最小堆的優先佇列，儲存序列的第K+1小元素到序列的最大元素。

壓入元素：

遍歷U陣列，確定當前A陣列中要壓入佇列的元素個數，分情況討論：

①q中元素不足k個

注意這裡不是直接將元素壓入q，而是先將元素壓入p，再從q裡取出最小值壓入q。

②q中元素大於等於k個

1’：當前要壓入的元素比q的最大值小，q的最大值壓入p後出隊，當前元素壓入q；

2’：當前要壓入的元素比q的最大值大（或等於），直接將當前元素壓入q。

輸出元素：

如果q陣列中元素個數小於k，依次將p的隊首元素壓入q，直到q陣列中元素個數等於k。

否則直接輸出q的隊首元素。

#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;//最小值優先
    priority_queue<int>q;//最大值優先
    int k=0,ac=0;//第k小、a陣列中的指標
    for(int i=0; i<n; ++i)//遍歷u陣列
    {
        ++k;
        int ps=p.size(),qs=q.size();
        for(int j=0; j<u[i]-ps-qs; ++j)//向佇列中壓入的a陣列的元素個數
        {
            if(q.size()<k)//q中元素不足k個
            {
                //先將元素壓入p，再從q裡取出最小值壓入q
                p.push(a[ac]);
                q.push(p.top());
                p.pop();
                ++ac;
            }
            else//q中元素大於k個
            {
                if(q.top()>a[ac])//當前要壓入的元素比q的最大值小
                {
                    //q的最大值壓入p後出隊，當前元素壓入q
                    p.push(q.top());
                    q.pop();
                    q.push(a[ac]);
                    ++ac;
                }
                else//當前要壓入的元素比q的最大值大
                {
                    //直接將當前元素壓入q
                    p.push(a[ac]);
                    ++ac;
                }
            }
        }
        if(q.size()<k)//如果q陣列中元素個數小於k
        {
            //依次將p的隊首元素壓入q，直到q陣列中元素個數等於k
            int temp=k-q.size();//需要壓入q的元素個數
            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
*/