Black Box平衡树

A - Black Box

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

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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

#include <cstdio>
#include <cstring>
#include <cstdlib>
#include<iostream>
using namespace std;
const int N = 30005;
const int MOD = 2147483647;
int a[N],u[N];

struct Node
{
     int left;      //左子树
     int right;     //右子树
     int size;      //大小
     int val;       //值
     int key;       //平衡种子
}tree[N];

int root,tot;

int add(int val){
    
    tot++;
    tree[tot].size = 1;
    tree[tot].val = val;
    tree[tot].key = rand();
    tree[tot].left = 0;
    tree[tot].right = 0;

    return tot;
}

void update_root(int now){
    
    int left,right;

    left = tree[now].left;
    right = tree[now].right;

    tree[now].size = tree[left].size + tree[right].size + 1;
    
}

//拆分(now原treap,a左子树,b右子树,val值)
void split_new(int now, int &a, int &b, int val){
    
    if(now == 0){
        a = 0;
        b = 0;
        return;
    }

    if(tree[now].val <= val){//now左子树中的所有值都小于now，
        a = now;
        split_new(tree[now].right, tree[a].right, b, val);
    }else{
        b = now;
        split_new(tree[now].left, a, tree[b].left, val);
    }

    update_root(now);
}//按权值aplit 
//
void merge_new(int &now, int a, int b){
    
    if(a==0 || b==0){
        now = a+b;
        return;
    }//某节点为空直接返回另一节点，合并结束 

    //按照key值合并(堆性质)
    if(tree[a].key<tree[b].key){
        /**
         * a树key值小于b树，那么b树属于a树的后代，因为b树恒大于a树，那么b树一定属于a树的右后代，
         * a的左子树不变，直接赋给now，递归合并a的右子树和b
         */
        now = a;
        merge_new(tree[now].right, tree[a].right, b);//b加到a的右后代上，a的右子树与y合并 
         
    }else{
        //a的key值大于b树，b的左子树与a合并 
        now = b;
        merge_new(tree[now].left, a, tree[b].left);
    }

    update_root(now);
}

void find_new(int now, int rank){//找第k大
    while (tree[tree[now].left].size+1 != rank){
        if(tree[tree[now].left].size >= rank){
            now = tree[now].left;
        }else{
            rank -= tree[tree[now].left].size + 1;
            now = tree[now].right;
        }
    }

    printf("%d\n", tree[now].val);
}

void insert_new(int val){
    int x,y,z;

    x = 0;
    y = 0;
    z = add(val);
    split_new(root, x, y, val);
    merge_new(x,x,z);
    merge_new(root, x, y);
}


void get_val(int rank){
    find_new(root, rank);
}

void solve(){
    int n,m,op,x,i;
    scanf("%d%d", &n,&m);

    memset(tree, 0, sizeof(tree));

    add(MOD);
    
    root = 1;
    tree[root].size = 0;
    for(i=0;i<n;i++){
        scanf("%d",&a[i]);
    }
    int k=0;
    for(int j=0;j<m;j++){
        
        scanf("%d",&x);
        for(;k<x;k++){
            insert_new(a[k]);
        }
        get_val(j+1);
    }
    
    
    
}

int main(){
    solve();
    return 0;
}