POJ1442 Black Box 优先队列（堆维护）+思维

1.题目描述：

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 12573		Accepted: 5154

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

2.题意概述：

当时题目看了好一会，不容易懂，大概意思是给定M个数，每次可以插入序列一个数；再给N个数，表示在插入第几个数时输出一个数，第一次输出序列中最小的，第二次输出序列中第二小的……以此类推，直到输出N个数。

对于样例：

7 4
3 1 -4 2 8 -1000 2
1 2 6 6
7代表下面给定7个数的数字序列，4可以理解为四次查询， 1 2 6 6为查讯，第一次查询是求数字序列只有前一个数（3）时，此时的第一小的数字，即1，第二次查询是求数字序列只有前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只有前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只有前6个数时，此时的第四小的数字。 

3.解题思路：

因为输出时是按照先输出最小的，再输出第二小这样的方式输出的，相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的，而是不断的在更新，所以用数组是无法实现的。我们可以用堆（优先队列）来做。
定义两个堆，一个用来存储前k小的数（大顶堆），大数在前，小数在后；另一个优先队列第k+1小到最大的数（小顶堆），小数在前，大数在后。每次拿到一个数，先判断第一个优先队列中的数满不满k个，如果不满k个，则直接把这个数压入到第一个队列；如果满k个，判断这个数和第一个优先队列中的第一个数的大小：如果比第一个数大，就压入第二个优先队列；如果比第一个数小，就把第一个优先队列的队首元素弹出压入第二个队列，把这个新数压入第一个优先队列。输出时，如果第一个优先队列里的元素个数小于k，则先把第二个优先队列里的队首元素弹出压入第一个优先队列，然后输出第一个优先队列的队首元素；如果满k个，则直接输出第一个优先队列的队首元素。

4.AC代码：

#include 
#include 
#include 
#include 
#include 
#define maxn 303000
using namespace std;
typedef long long ll;
ll num[maxn];
int n, m;
priority_queuebig;
priority_queue, greater >small;
int main()
{
	int m, n;
	scanf("%d%d", &m, &n);
	for (int i = 1; i <= m; i++)
		scanf("%lld", &num[i]);
	int u, cnt = 1;
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &u);
		while (cnt <= u)
		{
			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
			{
				ll n1 = big.top();
				ll n2 = small.top();
				big.pop();
				small.pop();
				big.push(n2);
				small.push(n1);
			}
			cnt++;
		}
		printf("%lld\n", small.top());
		big.push(small.top());//这句话很关键，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;
}