Black Box+POJ+treap树模板

Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 7437 Accepted: 3051

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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

解决方案：动态的添加数据，动态求第i大的数字。直接裸treap。

code：

<pre name="code" class="cpp">#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>using namespace std;struct node{    node *ch[2];    int r;///优先级    int v;///节点值    int s;    int cmp(int x)const    {        if(x==v) return -1;        else 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;    }///统计rank};void rotate(node* & rt,int d){    node *k=rt->ch[d^1];    rt->ch[d^1]=k->ch[d];    k->ch[d]=rt;    rt->maintain();    k->maintain();///必须先更新rt再更新k    rt=k;}///left rotate and right rotatevoid insert(node * & rt,int x){    if(rt==NULL)    {        rt=new node();        rt->ch[0]=rt->ch[1]=NULL;        rt->v=x;        rt->s=1;        rt->r=rand();///建立新的节点，插入数字    }    else    {        int d=(x<=rt->v?0:1);        insert(rt->ch[d],x);///选择合适的位置插入        if((rt->ch[d]->r)>rt->r)            rotate(rt,d^1);///左孩子优先级大右旋，右孩纸优先级大左旋    }    rt->maintain();///儿子节点发生改变，其父节点也要更新}int kth(node *rt,int k){    if(rt==NULL||k<=0||k>rt->s) return 0;    int s=(rt->ch[0]==NULL?0:rt->ch[0]->s);    if(s+1==k) return rt->v;    else if(s>=k) return kth(rt->ch[0],k);    else return kth(rt->ch[1],k-s-1);}///查询第k大的数字int v[30004];int main(){    int n,k;    while(~scanf("%d%d",&n,&k))    {        for(int i=1; i<=n; i++)        {            scanf("%d",&v[i]);        }        node *root=NULL;        int now=1;        for(int i=1; i<=k; i++)        {            int x;            scanf("%d",&x);            for(; now<=x; now++)                insert(root,v[now]);            printf("%d\n",kth(root,i));        }    }    return 0;}