[ACM] POJ 1442 Black Box (堆，优先队列）

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7099		Accepted: 2888

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

field=source&key=Northeastern+Europe+1996" style="text-decoration:none">Northeastern Europe 1996

解题思路：

题意不好懂。。

我直接解释下例子什么意思把。

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

7代表以下给定7个数的数字序列。4能够理解为四次查询， 1 2 6 6为查讯。第一次查询是求数字序列仅仅有前一个数（3）时。此时的第一小的数字，即1，第二次查询是求数字序列仅仅有前2个数（3 1）时。此时的第二小的数字,即 3，第三次查询是求数字序列仅仅有前6个数时（3,1，-4,2,8,-1000）。此时的第三小的数字，即1，第四次查询是求数字序列仅仅有前6个数时。此时的第四小的数字。

思路为： 当求第4小的数字时，我们用一个大顶堆来维护前三个最小的数字，那么小顶堆的堆顶即为所求。

用priority_queue<int>big;  优先队列还起到大顶堆的作用，顶部即为最大值  ，即 big.top();

    priority_queue<int,vector<int>,greater<int> >small;   小顶堆。顶堆为最小值 。即 small.top();

用例子来说明一下 大顶堆和小顶堆工作方法：

1 2 6 6

首先是1：  把第一个数3 插入到小顶堆中，这时候推断。假设大顶堆不为空且小顶堆的top小于大顶堆的top时，就把二者的top值互换，由于，大顶堆中的数不能比小顶堆中的大。大顶堆维护第i次查询时，前i-1个最小的数，这时候大顶堆为空，不用互换值。 输出第一次查询时前1个数的第1小的数即为 小顶堆的top　３，然后把小顶堆的top 移除。放到大顶堆中。

然后是2： 把第二个数1插入到小顶堆中。推断，大顶堆不为空，小顶.top() 1 <大顶.top() 3 ，所以二者互换，小顶.top()为3，大顶top()为1。 然后输出小顶.top()，即为第二小的数。把小顶的.top()移除，放入大顶中。这是大顶中维护的是眼下数字中前两个最小的数。

然后是6： 要求前6个数中第3小的数字，先得把3 1以后的四个数字插入才可以六个数。依次插入，  -4插入小顶堆。这时 大顶堆 3 1，-4<3,互换，  小顶堆 3，大顶堆为1 -4 ，2插入小顶堆，小顶堆 2,3  大顶堆 1 -4,  2>1,不用互换，  8插入到小顶堆。小顶堆为  2,3,8  大顶堆为 1。-4。  2>1,不用互换，-1000插入到小顶堆。小顶堆 -1000,2,3,8，大顶堆1，-4, -1000<1，不行，互换以后得小顶堆。1,2,3,8 ， 大顶堆 -1000。-4
，输出小顶堆.top() 1。 并把它移除放入到大顶堆中。为下次查询做准备。

所以从以上能够看出，关键点就是第i次查询时，大顶堆中维护的总是眼下数字中最小的 i-1个数。

代码：

#include <iostream>
#include <stdio.h>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn=30010;
int num[maxn];
int n,m;

int main()
{
    priority_queue<int>big;
    priority_queue<int,vector<int>,greater<int> >small;
    int m,n;
    scanf("%d%d",&m,&n);
    for(int i=1;i<=m;i++)
        scanf("%d",&num[i]);
    int cnt=1;
    int op;
    for(int i=1;i<=n;i++)
    {
        cin>>op;
        while(cnt<=op)
        {
            small.push(num[cnt]);
            if(!big.empty()&&small.top()<big.top())//小顶堆里面的数不能比大顶堆里面的数小
            {
                int n1=big.top();
                int n2=small.top();
                big.pop();
                small.pop();
                big.push(n2);
                small.push(n1);
            }
            cnt++;
        }
        printf("%d\n",small.top());
        big.push(small.top());//这句话非常关键。保证了在求第k个最小数时。大顶堆里面保存的是前k-1个最小数
        small.pop();
    }
    return 0;
}