poj 1442
|
Black Box
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
|
#include
#include
#include
using namespace std;
int main()
{
int c,n,m,x;
int a[300003];
while(~scanf("%d %d",&n,&m))
{
c=0;
priority_queue< int,vector,less >p;//大顶堆
priority_queue< int,vector, greater >q;//小顶堆一直维护大顶堆的最小
for(int i=0; iq.top())
{
int t=p.top();
p.pop();
p.push(q.top());
q.pop();
q.push(t);
}
printf("%d\n",q.top());
p.push(q.top());
q.pop();
}
}
return 0;
}
ps: 此部分是当时查题解时，找到的解释，感觉给了自己很大的助力
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；
怎么来做呢用的就是维护好两个堆，堆用优先队列来实现。
一个最大堆，一个最小堆。最大堆意思就是最大的元素在堆顶；最大堆存的是前i-1
小的数，那么最小堆的堆顶不就是要输出的第i小的数吗。
在这里我们用一个循环循环的作用就是每一次往最大堆离放i-1(或者i反正就是那么个
意思)个元素，代表前i-1小的数，那么我们循环里面再套一个while循环，因为你是在
1-u[p]这么多元素里找最小值，这些元素总得输入吧。每输入一个数据先放到q1(小)里
那么放进去之后堆顶元素就是这些数据当中最小的，如果最小的那个数比最大堆的堆顶
元素小，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。
那么在输出的时候同样，一个for循环结束后，最大堆的前i-1个最小的需要增至i个直接
把最小堆的堆顶元素赋值给最小堆最大堆堆顶弹出，