Poj1442 Black Box 二叉堆做法详解
Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 14267 | Accepted: 5831 |
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
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.
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
题目理解：每次输出给定数列中第k小的数，并且k的值递增，数列也在不断增加。
算法分析：
- 暴力解法：完全的每次读入后都进行排序，按照排序后的结果输出第k小的值，每次排序O(nlogn)，n次复杂度为O(n^2logn)。
- 稍微好一点的暴力做法：只用一个二叉堆，动态维护数列的最小值，复杂度O(logn)，虽然这里复杂度降低了，随着k的增大，每次需要pop掉和push进的数会越来越多（不断地pop来找到第k小，输出后还得把原来pop出来的数再push进去），所以复杂度仍然很高。
- 两个二叉堆，一个大根堆一个小根堆。作用为维护第k个值左边的数和右边的数，其中左边的数的最大值也不会超过右边的数的最小值，如果满足这个条件，就说明在一个单增的数列中，k这个位置的这个数字是合适的。由于每次k的值查询后都会递增1，所以每次查询后手动将k这个位置向右移动一位即可。具体看代码！！
#include
#include
#include
#include
using namespace std;
const int SIZE = 30010;
int a[SIZE],b[SIZE];
priority_queue , less > gq; //大根堆
priority_queue , greater > lq; //小根堆
int main()
{
int n,m;
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=m;i++)
scanf("%d",&b[i]);
int cnt=1;
for(int i=1;i<=m;i++)
{
while(cnt<=b[i])
{
lq.push(a[cnt]);
if(!gq.empty()&&gq.top()>lq.top())
//即：当且仅当插入的数小于原来第k小的数时（即当插入的数小于，原来第k小的数之前的大根堆中的最大值），第k小的数的具体数值才需要改变。比如-1,2,3中插入了4并不会影响第2小的数是2，但插入了-2后第2小的数会变为-1.
{
int t=gq.top();
int t1=lq.top();
gq.pop();lq.pop();
gq.push(t1);lq.push(t);
}
cnt++;
}
printf("%d\n",lq.top());
int t=lq.top();
lq.pop();
gq.push(t);//手动向右移动一位，相当于查询后k自增1。
}
return 0;
}