POJ 1442 （treap）

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 9871		Accepted: 4043

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

题意：题意还是太长了，其实还是很好懂的题意，简单的说就是问你，动态查询有x个数字的数组，他第i大的数字是多少？这

题解：这里我们很容易想到使用set，map之类的，但是我想了一下，好像set不能直接查询第k大的数字呀！只能一次一的遍历呀，于是查了资料确实是不能查询，好吧查了下题解，看到2种做法，一是使用2个优先队列维护第k大的关系，第二个正是我要说的treap，treap封装了类似set，map的排序二叉树的方式维护数字的大小，除此之外，他还可以快速查找第k大的元素，查询数字x的名次，真是太方便啦！那么这一题就是treap的模板题了，这里我们的kth函数有2种查询方式，较小优先和较大优先。和set的greater和less是不是很像？

#include   
#include
#include   
#include
#include
using namespace std;
struct Node
{
	Node *ch[2];
	int r;
	int v;
	int s;
	Node() {}
	Node(int v) : v(v) {
		ch[0] = ch[1] = NULL; r = rand(); s = 1;
	}
	bool operator < (const Node& rhs) const {
		return r < rhs.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;
	}
};

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(x);
	}
	else {
		int d = (x < o->v ? 0 : 1);
		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* u = o;
		if (o->ch[0] != NULL && o->ch[1] != NULL) {
			int d2 = (o->ch[0] > o->ch[1] ? 1 : 0);
			rotate(o, d2); remove(o->ch[d2], x);
		}
		else {
			if (o->ch[0] == NULL) o = o->ch[1];
			else o = o->ch[0];
			delete u;
		}
	}
	else
		remove(o->ch[d], x);
	if (o != NULL) o->maintain();
}

int kth_small(Node* o, int k) //数字小优先
{
	if (o == NULL || k <= 0 || k > o->s)
		return 0;
	int s = (o->ch[0] == NULL ? 0 : o->ch[0]->s);
	if (k == s + 1) return o->v;
	else if (k <= s) return kth_small(o->ch[0], k);
	else return kth_small(o->ch[1], k - s - 1);
}

int kth_big(Node *o, int k)//数字大优先
{
	if (o == NULL || k <= 0 || k > o->s)
		return 0;
	int s = o->ch[1] == NULL ? 0 : o->ch[1]->s;
	if (s + 1 == k)
		return o->v;
	else {
		if (s >= k)
			return kth_big(o->ch[1], k);
		else
			return kth_big(o->ch[0], k - s - 1);
	}
}

void remove_tree(Node *&o) {
	if (o->ch[0] != NULL)
		remove_tree(o->ch[0]);
	if (o->ch[1] != NULL)
		remove_tree(o->ch[1]);
	delete o;
	o = NULL;
}
int n, m, a[30010];
Node *rt = NULL;
int main()
{
#ifdef CDZSC  
	freopen("i.txt", "r", stdin);
#endif  
	while (scanf("%d %d", &n, &m) != EOF)
	{
		srand(time(0));
		for (int i = 1; i <= n; i++)
			scanf("%d", &a[i]);
		int l = 1;
		for (int i = 1; i <= m; i++)
		{
			int x;
			scanf("%d", &x);
			while (l <= x)
			{
				insert(rt, a[l]);
				l++;
			}
			//for (int j = 1;j<=i;j++)
			printf("%d\n", kth_small(rt, i));
			//printf(" %d\n", kth_big(rt, i));
		}
		remove_tree(rt);
	}
	return 0;
}