POJ_P1442 Black Box(Treap模板题+动态第k小)

传送门
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 9588 Accepted: 3921
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 Blak 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

题目大意
给你一组数，询问插入第x个数时第i小的数

#include
#include
#include
#include
using namespace std;
struct Node{
    Node *ch[2];//左右子树 
    int r,v,s; //随机优先值，值，节点总数 
    Node(int v):v(v){ch[0]=ch[1]=NULL;r=rand();s=1;}
    bool operator < (const Node &a) const{return rint comp(int x) const{if(x==v) return -1;return x0:1;}
    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){//旋转 d=0左旋 d=1右旋 
    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){//插入x 
    if(o==NULL) o=new Node(x);
    else{
        int d=(xv?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){//删除节点 
    int d=o->comp(x);
    if(d==-1){
        Node* u=o;
        if(o->ch[0]!=NULL&&o->ch[1]!=NULL){
            int d2=(o->ch[0]->r > o->ch[1]->r?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(Node* o,int k){//第k小 
    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(o->ch[0],k);
    else return kth(o->ch[1],k-s-1); 
}

#define N 30005 
int n,m,x,done;int v[N];

int main(){
    scanf("%d%d",&n,&m);root=NULL;
    for(int i=1;i<=n;i++) scanf("%d",&v[i]);
    for(int i=1;i<=m;i++){
        scanf("%d",&x);
        while(doneprintf("%d\n",kth(root,i));
    }
    return 0;
}