Black Box
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 9883 | Accepted: 4047 |
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
1A~果然上午看小说，中午吃好吃的，下午效率高23333
说题意，有两种操作 add和query，query每次查询的是第i小的，i是从0开始的，查询之前i++，查询之后不删除。告诉你add数字的依次次序，又告诉你当数组当前有几个数的时候进行了一次查询（这句话我结合示例看了好久才明白什么意思）然后就是小case了
/*************
poj1442
2016.1.26
984K 204MS C++ 2246B
*************/
#include <cstdio>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
#define max(x,y) ((x)>(y)?(x):(y))
#define min(x,y) ((x)>(y)?(y):(x))
#define INF 0x3f3f3f3f
#define MAXN 100005
using namespace std;
int cnt=1,rt=0; //节点编号从1开始
struct Tree
{
int key, size, pri, son[2]; //保证父亲的pri大于儿子的pri
void set(int x, int y, int z)
{
key=x;
pri=y;
size=z;
son[0]=son[1]=0;
}
}T[MAXN];
void rotate(int p, int &x)
{
int y=T[x].son[!p];
T[x].size=T[x].size-T[y].size+T[T[y].son[p]].size;
T[x].son[!p]=T[y].son[p];
T[y].size=T[y].size-T[T[y].son[p]].size+T[x].size;
T[y].son[p]=x;
x=y;
}
void ins(int key, int &x)
{
if(x == 0)
T[x = cnt++].set(key, rand(), 1);
else
{
T[x].size++;
int p=key < T[x].key;
ins(key, T[x].son[!p]);
if(T[x].pri < T[T[x].son[!p]].pri)
rotate(p, x);
}
}
void del(int key, int &x) //删除值为key的节点
{
if(T[x].key == key)
{
if(T[x].son[0] && T[x].son[1])
{
int p=T[T[x].son[0]].pri > T[T[x].son[1]].pri;
rotate(p, x);
del(key, T[x].son[p]);
}
else
{
if(!T[x].son[0])
x=T[x].son[1];
else
x=T[x].son[0];
}
}
else
{
T[x].size--;
int p=T[x].key > key;
del(key, T[x].son[!p]);
}
}
int find(int p, int x) //找出第p小的节点的编号
{
if(p == T[T[x].son[0]].size+1)
return x;
if(p > T[T[x].son[0]].size+1)
find(p-T[T[x].son[0]].size-1, T[x].son[1]);
else
find(p, T[x].son[0]);
}
int ad[30005],qu[30005];
int main()
{
// freopen("cin.txt","r",stdin);
int m,n,pos,posq;
while(~scanf("%d%d",&m,&n))
{
rt=0;
cnt=1;
for(int i=1;i<=m;i++) scanf("%d",&ad[i]);
for(int i=1;i<=n;i++) scanf("%d",&qu[i]);
pos=0,posq=1;
for(int i=1;i<=m;i++)
{
ins(ad[i],rt);
// cout<<"1111111"<<endl;
while(i==qu[posq])
{
// cout<<"22222222"<<endl;
pos++;
int p=find(pos,rt);
printf("%d\n",T[p].key);
posq++;
}
}
}
return 0;
}
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