Black Box POJ1442

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

Source

Northeastern Europe 1996

题意：

给一系列数字，给出前k个数字中第i大的数字，i从1->m;

题解：

维护一个从大到小的优先队列和一个从小到大的，大顶堆中存放的是前i-1个数字的最小值，但是会时时更新，小顶堆中存放的是当前的最小值

此处记录一下改变优先队列的大小顺序的方法，

1，首先优先队列默认从大到小，大的在顶

2，从小到大。

priority_queue<int,vector<int>,greater<int> >//这样便是从小到大

priority_queue< int,vector<int>,less<int> > //大->小

3，如果是结构体

struct number1
{
    int x;
    bool operator < (const number1 &a) const//只有 < 这个符号
    {
        return x>a.x;//小值优先    //反之大值优先
    }
};

　　

　　

//#include <bits/stdc++.h>
#include <cstdio>
#include <queue>
using namespace std;
const int MAXN=30010;
priority_queue<int>big;
priority_queue<int,vector<int>,greater<int> >mi;

int a[MAXN],b[MAXN];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for (int i = 0; i <n ; ++i) {
        scanf("%d",&a[i]);
    }
    int op;
    int k=0;
    for (int i = 0; i <m ; ++i) {
        scanf("%d",&op);
        while(k<op)
        {
            mi.push(a[k]);
            if(!big.empty()&&mi.top()<big.top())//大顶堆维护前k-1的最小值，小顶堆维护当前除了前k-1个最小值的最小值。
            {
                int t;
                t=big.top();
                big.pop();
                big.push(mi.top());
                mi.pop();
                mi.push(t);
            }
            k++;
        }
        printf("%d
",mi.top());//需要把当前的最小值发放入到大顶堆中
        big.push(mi.top());
        mi.pop();
    }

    return 0;
}