POJ 1442 Blackbox

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

问题分析与解题思路

题意

输入一行数A[];再输入一行数u[]；让你输出前[]的前u[p]个数据中，第p+1小的数，正如给出的例子

7 4
3 1 -4 2 8 -1000 2
1 2 6 6
前1个数中第1小的数是3；
前2个数中第2小的数是3；
前6个数中第3小的数是1；
前6个数中第4小的数是2；

通过维护一个最大堆，一个最小堆进行求解。

• 最大堆中存放第1到第i-1小的数
• 最小堆中存放第i-1到第n小的数（n为当前数字总数）

所以最小堆的堆顶就是我们要的结果。

每插入一个数，我们先把它放入最小堆，更新最小堆，如果最小堆堆顶元素小于最大堆堆顶元素，说明前i-1个最小的数据需要更新，只要将小堆堆顶和大堆堆顶元素交换即可。再把最小堆的堆顶元素弹出输出，然后赋值给最大堆堆顶。（因为下一次所求不是第i小而是第i+1小的元素）

数据结构与算法设计及其主要代码段

定义堆

priority_queue<int> q2;  //最大堆(默认)；
priority_queue<int,vector<int>,greater<int> > q1;//最小堆(自定义)

插入与get

for(i=0;i<m;i++)
    {
        scanf("%d",&k);
        while(j<k)
        {
            q1.push(A[j]);
            // 最小堆堆顶元素小于最大堆堆顶元素
            if(!q2.empty() && q1.top()<q2.top())
            {
                int t;
                t=q1.top();
                q1.pop();
                q1.push(q2.top());
                q2.pop();
                q2.push(t);
            }
            j++;
        }
        // 碰到get弹出输出最小堆堆顶，并加入最大堆
        printf("%d\n",q1.top());
        q2.push(q1.top());
        q1.pop();
    }

程序运行结果及分析

A. 算法复杂度

• 建堆:O(n)
• 插入删除：O(logn)
• 总的复杂度:O(n)

B. 运行时间

内存:2260kB, 时间:52ms（数据来自openjudge）

心得体会与总结

1. 使用stl的priority_queue会让本题难度降低很多。
2. 本题最重要的就是想清楚两个堆中所存的元素，最大堆中存放第1到第i-1小的数，最小堆中存放第i-1到第n小的数（n为当前数字总数。
3. 本题巧妙之处在于，一般堆都是用来解决最大，最小问题，两个堆的使用可以解决第k大／小问题。