Black Box－－［优先队列 最大堆最小堆的应用］

Posted 2020-06-23 kiraa

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

解题思路分析：
拿到这道题最先想到的思路是（错误）：
　　1.用队列保存add命令，用vector保存进入黑箱的元素。
　　2.每次向黑箱加入元素后，对黑箱中元素排序，输出第i个元素。
这个算法看似没用问题，但实际上，经过极端数据测试，在M、N均为30000时，跑完程序需要2.7s，会超时，原因是每一次添加完元素后都要对所有元素排序，时间开销太大了。然而经过分析容易发现，实际上并没有必要每次都对所有元素排序，我们只需要保存前i－1小的元素，然后找出第i个元素起之后的最小元素与前i－1个元素中最大的元素比较，即可得到第i小的元素。这样思路就明了了，
下面给出正确思路：
　　1.用优先队列分别建立最小堆（顶部最小）和最大堆（顶部最大）；
　　2.每次i增加时，将最小堆顶部元素加入最大堆；
　　3.每次加入元素时比较最小堆顶部元素和最大堆顶部元素，如果“top小”>“top大”，则交换两个堆顶部元素；
　　4.输出最小堆顶部元素。
注意优先队列的两种用法：

　1. 标准库默认使用元素类型的<操作符来确定它们之间的优先级关系。

priority_queue<int> q;

　通过<操作符可知在整数中元素大的优先级高。
 　2. 数据越小，优先级越高  greater<int> 定义在头文件 <functional>中

priority_queue<int, vector<int>, greater<int> >q; 

3.自定义比较运算符<

代码如下：

 1 #include <iostream>
 2 #include <set>
 3 #include <vector>
 4 #include <cstdio>
 5 #include <queue>
 6 #include <algorithm>
 7 #include <time.h>
 8 #include <functional>
 9 using namespace std;
10 #define clock__ (double(clock())/CLOCKS_PER_SEC)
11 
12 int M,N;
13 queue<int> A;//存储add命令
14 priority_queue<int,vector<int>,greater<int> > A_min;//最小堆
15 priority_queue<int> A_max;//最大堆
16 int i;
17 
18 int main() {
19
20     i=0;
21     scanf("%d%d",&M,&N);
22     while (M--) {
23         int a;
24         scanf("%d",&a);
25         A.push(a);
26     }
27     while(N--){
28         int u;
29         scanf("%d",&u);
30         i++;
31         if(A_min.size()!=0) {
32             A_max.push(A_min.top());
33             A_min.pop();
34         }
35         int n=u-(A_min.size()+A_max.size());
36         while(n--){
37             A_min.push(A.front());A.pop();
38             if(A_max.size()&&A_min.size()&&A_min.top()<A_max.top()){
39                 int a=A_max.top(); A_max.pop();
40                 int b=A_min.top(); A_min.pop();
41                 A_min.push(a);A_max.push(b);
42             }
43         }
44         printf("%d\n",A_min.top());
45         
46     }
47    // cout<<"time : "<<clock__<<endl;
48     return 0;
49 }