POJ 1442 Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8627		Accepted: 3537

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

题目大意：输入n，m 下一行给你n个数A1...An依次进行插入排序，然后给你m个数 u1...um要求你输出进行第ui次插入时候，序列中第i小的数是多少

思路使用两个优先队列 一个从小到大，一个从大到小，从小到大的队列Now储存第i+1大到第ui大的数，另外一个从大到小队列Del储存最小到第i小的数，每次插入操作的时候将A[i]插入到Del中然后将Del的队顶插入到Now中

例如 第5次插入前

Del ： 1 ,-4 Now：2, 3

之后插入 8

Del：8,1,-4 Now：2,3

Del的队头出队到Now中

Del 1,-4 Now：2,3,8

如果这次要查询的话直接输出Now的队头2然后将2插入到Del的队头就可以了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<queue>
#include<algorithm>
using namespace std;
#define inf 0xffffff
#define maxn 30010

int A[maxn],u[maxn];

int main()
{
int n, m;
while(cin>>n>>m)
{
for(int i = 1; i <= n; i++)
scanf("%d",&A[i]);
for(int i = 0; i < m; i++)
scanf("%d",&u[i]);
int top = 0;

priority_queue<int,vector<int>,greater<int> > Now;
priority_queue<int> Del;
while(!Now.empty()) Now.pop();
while(!Del.empty()) Del.pop();

for(int i = 1; i <= n; i++)
{
if(Del.empty() || A[i] >= Del.top()) //一个优化当Del为空就代表还没有查询或者
Now.push(A[i]);<span style="white-space:pre">			</span> //<span style="font-family: 'Times New Roman', Times, serif;">插入数据大于第i小的数可以直接插入节约20ms</span>
else
{
Del.push(A[i]);
int x = Del.top();
Del.pop();
Now.push(x);
}
while(i == u[top] && top < m)
{
int x = Now.top();
Now.pop();
printf("%d\n",x);
Del.push(x);
top++;
}
}
}
return 0;
}