POJ 1442 Treap 解题报告

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

3
3
1
2

【解题报告】
题意： 给n个数，m个询问，对第i数个询问前Xi个数中第i小的是那个数。
Treap直接上。

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 100010

int n,m,a[N];
int q_rand()
{
    static int seed=10007; 
    return seed=int(seed*48271LL%2147483647);
}
typedef struct Node
{
    Node *l,*r;
    int val,pri;
    int sz;
    Node(int x)
    {
        l=r=NULL;
        pri=q_rand();
        val=x;
        sz=1;
    }
}Node;
Node *root;

int getsz(Node *T)
{
    return(T==NULL)?0:T->sz;
}
Node *L_rotate(Node *T)
{
    Node *A=T->r;
    T->r=A->l;
    A->l=T;
    A->sz=T->sz;
    T->sz=getsz(T->l)+getsz(T->r)+1;
    return A;
}
Node *R_rotate(Node *T)
{
    Node *A=T->l;
    T->l=A->r;
    A->r=T;
    A->sz=T->sz;
    T->sz=getsz(T->l)+getsz(T->r)+1;
    return A;
}
void insert(Node *&T,int val)
{
    if(T==NULL) {T=new Node(val);return;}
    T->sz++;
    if(T->val>=val)
    {
        insert(T->l,val);
        if((T->l->pri)<(T->pri)) T=R_rotate(T);
    }
    else
    {
        insert(T->r,val);
        if((T->r->pri)<(T->pri)) T=L_rotate(T);
    }
}
int find(Node *T,int k)
{
    int tmp=getsz(T->l)+1;
    if(tmp==k) return T->val;
    if(tmp>k) return find(T->l,k);
    if(tmp<k) return find(T->r,k-tmp);
}
int main()
{
    root=NULL;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;++i) scanf("%d",&a[i]);
    int j=1;
    for(int i=1;i<=m;++i)
    {
        int x;scanf("%d",&x);
        while(j<=x)
        {
            insert(root,a[j]);
            ++j;
        }
        printf("%d\n",find(root,i));
    }
    return 0;
}