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 }