I, entitled
Black Box represents a primitive database.
It can be used to store an array of integers, and it has a special variable i.
In the beginning, the black box is empty, and i = 0.
Now black box series operation process, comprising the following two operations:
1, ADD (x): x represents added to the black box.
2, GET: i is incremented by 1, the output of the black box a small value (ie, all the i-th number of press after ascending order) of the i.
A specific example is given below:
序号 操作 i 盒子内数（升序排列后） 输出的值
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
For convenience of description, we define the following two sequences:
1, A (1), A (2), ..., A (M): This sequence is obtained from the addition of all the elements are arranged in a black box after the order of addition, A in the above example of the sequence (3,1, - 4,2,8, -1000,2).
2, u (1), u (2), ..., u (N): This item i is represented by the sequence number of a GET operation within the i-th element in the box. u sequence example is (1,2,6,6).
Now they're all values obtained during the operation in accordance with the output given sequences A and u.
Input Format
Inputs include three lines.
The first line contains two integers M and N, the length of sequence A and sequence u.
The second line contains M integers, representing each element A sequence.
The third line contains N integers, each element u represents the sequence.
Separated by a space between each number of peers.
Output Format
All numerical GET operation output during the output operation.
Each value per line.
data range
| A (I) | <2 * 109 = | A (I) | <= 2 * 109,
1≤N≤M≤300001≤N≤M≤30000,
all pp (1≤p≤N1≤p≤N ), p≤u (p) ≤Mp≤u ( p) ≤M established
Sample input:
7 4
3 1 -4 2 8 -1000 2
1 2 6 6
Sample output:
3
3
1
2
| Difficulty: Moderate |
| Time / space restrictions: 1s / 64MB |
| Total by: 113 |
| The total number of attempts: 193 |
| Source: "Advanced algorithm Competition Guide" |
Second, the idea
This question is examined concept large pile top, by maintaining a small sequence data stack top and the large stack top (small or large heap root root pile), in the maintenance process, the search order is to pay attention to:
/ *
Steps of the algorithm ideas:
The first step: from the perspective of number theory to moderation, taking the form of its data, relationships, recursive formula.
Step two: Starting from the data structure point designed algorithm explanations steps, generally having a data structure and operation of the coupling (i.e., an algorithm structure and algorithm operating a correspondence relationship).
The third step: from programming, the realization of the program.
* /
/*
Sequential search algorithm: enumeration does not leak; storage state variables
*/
Third, to achieve
/*算法搜索中的三个问题：1.按什么样的顺序搜索2.怎样枚举才能不重不漏所有情况或变量3.变量或状态用数组还是多个变量存储*/
//算法的一般思路:考察对顶推，用两个堆；来维护一个序列。
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
const int N=30010;
int n,m; //n表示ADD操作次数，m表示get操作次数
int a[N],b[N];
int main()
{
cin>>n>>m;
for(int i=0;i<n;i++)cin>>a[i];
for(int j=0;j<m;j++)cin>>b[j];
sort(b,b+m);
priority_queue<int> left;
priority_queue<int,vector<int>,greater<int>> right;
/*题解思想：通过维护大顶堆、小顶堆，完成这个模拟操作操作*/
//b[m]数组暗含了add(x)和get两个操作
int i=0,j=0; //枚举操作确定索引变量
//程序设计
while(i<n || j<m) //设计：一步一步操作
{
while(j<m && b[j]==i)
{
//get操作 在i个元素情况下，get的操作。
cout<<right.top()<<endl;
left.push(right.top());
right.pop();
j++;
}
//add操作分两种情况
int x=a[i];
if(left.empty()||x>=right.top()) right.push(x);
else
{
left.push(x);
right.push(left.top());
left.pop();
}
i++;//i表示数组a的索引号
}
return 0;
}
