N Transaction i Black Box contents after transaction Answer

      (elements 升序排列)

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   

7 4
3 1 -4 2 8 -1000 2
1 2 6 6

3
3
1
2

此题可用二叉搜索树解决。

二叉搜索树的建立：每拿到一个元素，比根小放在左根，比根大放在右根，否则放根上。

t.weight 表示结点上有几个数, t.key表示数的值 ,t.size表示以该点为根的树上有多少数

重点是左旋和右旋！

2个注意：注意把a的父亲结点改掉，注意把a,b的父亲结点改掉

下图转载自http://dongxicheng.org/structure/treap/

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<functional>
using namespace std;
#define MAXM (30000+10)
#define MAXN (30000+10)
struct tree_node
{
	int key,fim,size,weight;
	tree_node *left,*right,*father;
	int l_size()
	{
		return (left==NULL)?0:(left->size);
	}
	int r_size()
	{
		return (right==NULL)?0:(right->size);
	}
	void count_size()
	{
		//size=(left==NULL)?0:(left->size)+(right==NULL)?0:(right->size)+weight;
		size=l_size()+r_size()+weight;
	}
	tree_node():key(0),fim(rand()),size(0),weight(0){left=right=father=NULL;}
	tree_node(int _key):key(_key),fim(rand()),size(1),weight(1){left=right=father=NULL;}
};
tree_node* newnode(int key)
{
	tree_node* p;
	p=new tree_node;
	*p=tree_node(key);
	return p;
}
struct Treep
{
	tree_node *root;
	Treep(){root=NULL;}
	void left_rotaue(tree_node *&a)
	{
		tree_node* b=a->right;
		a->right=b->left;if (a->right!=NULL) a->right->father=a;
		b->left=a;b->father=a->father;a->father=b;
		a->count_size();
		b->count_size();
		if (b->father)
		{
			if (b->father->left==a) b->father->left=b;
			else b->father->right=b;
		}
		a=b;
	}
	void right_rotaue(tree_node *&a)
	{
		tree_node* b=a->left;
		a->left=b->right;if (a->left!=NULL) a->left->father=a;
		b->right=a;
		b->father=a->father;
		a->father=b;
		a->count_size();
		b->count_size();
		if (b->father)
		{
			if (b->father->left==a) b->father->left=b;
			else b->father->right=b;
		}
		a=b;
	}
	void insert(int key)
	{
		if (root==NULL) {root=newnode(key);return;}
		tree_node *now=root;
		while (1)
		{
			now->size++;
			if (key==now->key) {now->weight++;break;}
			else if (key<now->key)
			{
				if (now->left==NULL) {now->left=newnode(key);now->left->father=now; now=now->left; break;}
				else now=now->left;
			}
			else
			{
				if (now->right==NULL) {now->right=newnode(key);now->right->father=now; now=now->right; break;}
				else now=now->right;
			}
		}
		for (;now!=root&&now->father->fim<now->fim;)
		{
			if (now->key<now->father->key) {now=now->father; right_rotaue(now);}
			else {now=now->father; left_rotaue(now); }
			if (now->father==NULL) root=now;
		}

	}
	int gets(int k)
	{
		tree_node* now=root;
		while (1)
		{
			if (now->l_size()>=k) {now=now->left;}
			else if (now->l_size()+now->weight>=k) return now->key;
			else {k-=now->l_size()+now->weight; now=now->right;}
		}
	}
	void print(tree_node *now)
	{
		if (now->left!=NULL)
		{
			cout<<"( ";
			print(now->left);
			cout<<" )";
		}
		cout<<" "<<now->key<<" ";
		if (now->right!=NULL)
		{
			cout<<"( ";
			print(now->right);
			cout<<" )";
		}


	}
}t;
int n,m,a[MAXN],u[MAXM];
int main()
{
//	freopen("poj1442.in","r",stdin);
	cin>>n>>m;
	for (int i=1;i<=n;i++) cin>>a[i];
	for (int i=1;i<=m;i++) cin>>u[i];
	for (int i=1,j=1;i<=n;i++)
	{
		t.insert(a[i]);//t.print(t.root);cout<<endl;
		while (j<=m&&i==u[j]) {cout<<t.gets(j)<<endl;j++;}
	}
	return 0;
}