题目描述
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:
-
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).
-
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§ <= 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§) sequence.
输入
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.
输出
Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.
思路
这道题就是让我们求一个在线数列，对于每个询问
，
,求当数列元素个数为
,时第
小的数。
怎么处理这种情况呢？
我们可以开两个堆:
大根堆
小根堆
对于每个询问
，经过操作，
输出堆顶即可。
操作：
- 把还没进队列的数推进
- 维护一下两个堆，使 所有元素小于 最小(堆顶)元素.
- 此时，输出结果后，将 最小元素推进 中，此时， 的元素个数保持 (操作完后)个。
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cstdlib>
#include<cmath>
using namespace std;
const int N=3e4+10;
struct heap_max
{
int a[N],len;
heap_max(){};
inline void m_swap(int i,int j){int t=a[i];a[i]=a[j];a[j]=t;}
inline void up(int j)
{
while(j!=1)
{
int i=j>>1;
if(a[i]<a[j])m_swap(i,j),j=i;
else break;
}
}
inline void down(int i)
{
int j=i<<1;
while(j<=len)
{
if(j<len&&a[j+1]>a[j])j++;
if(a[i]<a[j])m_swap(i,j),i=j,j=i<<1;
else break;
}
}
void ins(int x){a[++len]=x;up(len);}
inline int top(){return a[1];}
void pop(){a[1]=a[len--];down(1);}
}in;
struct heap_min
{
int a[N],len;
heap_min(){};
inline void m_swap(int i,int j){int t=a[i];a[i]=a[j];a[j]=t;}
inline void up(int j)
{
while(j!=1)
{
int i=j>>1;
if(a[i]>a[j])m_swap(i,j),j=i;
else break;
}
}
inline void down(int i)
{
int j=i<<1;
while(j<=len)
{
if(j<len&&a[j+1]<a[j])j++;
if(a[i]>a[j])m_swap(i,j),i=j,j=i<<1;
else break;
}
}
void ins(int x){a[++len]=x;up(len);}
inline int top(){return a[1];}
void pop(){a[1]=a[len--];down(1);}
}out;
int query[N],a[N];
int main()
{
int n,m,len=1;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",&query[i]);
for(int i=1;i<=m;i++)
{
while(len<=query[i])out.ins(a[len++]);
while(out.top()<in.top()&&in.len)
{
int x=out.top(),y=in.top();
in.pop();out.pop();
in.ins(x);out.ins(y);
}
printf("%d\n",out.top());
in.ins(out.top());
out.pop();
}
return 0;
}
``