POJ1442--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

3312

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
#define maxn 130080
int A[maxn],B[maxn];
struct Node
{
	Node * ch[2];
	int r,v,s;
	bool operator < (const Node & a) const
	{
		return r < a.r;
	}
	int cmp(int x) const
	{
		//if(x == v)	return  -1;
		return x < v?0:1;
	}
	void maintain()
	{
		s = 1;
		if(ch[0] != NULL)	s += ch[0] -> s;
		if(ch[1] != NULL)	s += ch[1] -> s;
	}
}*root;

void rotate(Node *&o,int d)
{
	Node * k = o -> ch[d^1];
	o -> ch[d^1] = k -> ch[d];
	k -> ch[d] = o;
	o -> maintain();
	k -> maintain();
	o = k;
}

void insert(Node *& o,int x)
{
	if(o == NULL)	
	{
		o = new Node();
		o -> ch[0] = o -> ch[1] = NULL;
		o -> v = x;
		o -> r = rand();
		o -> s = 1;
	}
	else
	{
		int d = o -> cmp(x);
		insert(o -> ch[d],x);
		if(o -> ch[d] -> r > o -> r)	rotate(o,d^1);
	}
	o -> maintain();
}

void remove(Node *& o,int x)
{
	int d = o -> cmp(x);
	if(d == -1)
	{
		Node * tmp = o;
		if(o -> ch[0] == NULL)	
		{
			o = o -> ch[1];
			delete tmp;
			tmp = NULL;
		}
		else if(o -> ch[1] == NULL)	
		{
			o = o -> ch[0];
			delete tmp;
			tmp = NULL;
		}
		else 
		{
			int d2 = (o -> ch[0] -> r > o -> ch[1] -> r?1:0);
			rotate(o,d2);
			remove(o -> ch[d2],x);
		}
	}
	else remove(o -> ch[d],x);
	if(o != NULL)	o -> maintain();
}

bool find(Node * o,int x)
{
	while(o != NULL)
	{
		int d = o -> cmp(x);
		if(d == -1)	return 1;
		else o = o -> ch[d];
	}
	return 0;
}


int Rank(Node * o,int x)
{
	int ans = 0;
	while(o)
	{
		if(o -> v == x)
		{
			if(o -> ch[0])	ans += o -> ch[0] -> s;
			return	ans;
		}
		else if(o -> v > x)
		{
			o = o -> ch[0];
		}
		else 
		{
			if(o -> ch[0])	ans += o -> ch[0] -> s;
			ans++;
			o = o -> ch[1];
		}
	}
	return ans;
}

int Kth(Node *o,int k)//这里是要返回第k小的元素
{
	
	while(k)
	{
		if(o -> ch[0])
		{
			if(o -> ch[0] -> s >= k)	o = o -> ch[0];
			else if( o -> ch[0] -> s == k-1)	return o -> v;
			else k -= o-> ch[0] -> s + 1,o = o -> ch[1];
		}
		else if( k == 1 )	return  o -> v;
		else k--,o = o -> ch[1];
	}
	/*
	int  tmp = 0;
	if(o -> ch[0])
	{
		tmp += o -> ch[0] -> s;
	}
	if(k == tmp + 1)	return o -> v;
	else if(k <= tmp)	return Kth(o -> ch[0],k);
	else return Kth(o -> ch[1],k - 1 - tmp);
	*/
}

void DeleteTreap(Node *& o)
{
	if(o == NULL)	return;
	if(o -> ch[0] != NULL)	DeleteTreap(o -> ch[0]);
	if(o -> ch[1] != NULL)	DeleteTreap(o -> ch[1]);
	delete o;
	o = NULL;
}

int main()
{
//	freopen("in.txt","r",stdin);
	int n,m;
	while(scanf("%d%d",&n,&m)==2)
	{
		int now = 1;
		for(int i = 1;i <= n;i++)	scanf("%d",&A[i]);
		for(int i = 1;i <= m;i++)
		{
			int x;
			scanf("%d",&x);
			for(;now <= x;now++)
			{
				insert(root,A[now]);
			}
			printf("%d\n",Kth(root,i));
		}
	}
	return 0;
}