POJ 1442-Black Box(优先队列)

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

题意 ：给出n，m，然后接下一行输入n个数，最后一行输入m个数，要求对于最后一行的第i个数，输出在原数列中前x个数中第i小的数，。好绕口，然后一开始用sort是各种TLE，好多大神用平衡树过的，差点没吓哭我，然后，。。撸了两个堆，一个最大堆，一个最小堆，注意维护好这两个堆，其中最大堆放前k个最小的数，最小堆放剩余的数，那么最大堆的堆顶就是第k个最小的数。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <algorithm>
#include <set>
#include <vector>
#include <string>
#include <map>
#include <queue>
using namespace std;
const int maxn= 30010;
int a[maxn];
priority_queue <int,vector<int>,less<int> > maxQ;
priority_queue <int,vector<int>,greater<int> > minQ;
int main()
{
	int n,m,x;
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++)
        scanf("%d",&a[i]);
	int pos=0;
	for(int i=1;i<=m;i++)
	{
		scanf("%d",&x);
		while(pos<x)
		minQ.push(a[pos++]);//新加元素放入最小堆
		while(maxQ.size()<i)//维护最大堆
		{
			maxQ.push(minQ.top());
			minQ.pop();
		}
		while(!minQ.empty()&&maxQ.top()>minQ.top())//维护两个堆，严格保证maxQ.top()<=minQ.top()
		{
			int t1=maxQ.top(),t2=minQ.top();
			maxQ.pop();minQ.pop();
			maxQ.push(t2);minQ.push(t1);
		}
		printf("%d\n",maxQ.top());
	}
	return 0;
}