POJ 1442 动态第k大

Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 6365 Accepted: 2580

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

Source

Northeastern Europe 1996

treap入门学习，直接上代码：

/* ***********************************************Author :xianxingwuguanCreated Time :2014/3/1 14:54:52File Name :5.cpp************************************************ */#pragma comment(linker, "/STACK:102400000,102400000")#include <stdio.h>#include <iostream>#include <algorithm>#include <sstream>#include <stdlib.h>#include <string.h>#include <limits.h>#include <string>#include <time.h>#include <math.h>#include <queue>#include <stack>#include <set>#include <map>using namespace std;#define INF 0x3f3f3f3f#define eps 1e-8#define pi acos(-1.0)typedef long long ll;const int maxn=100100;struct Treap{int root,treapnode,key[maxn*5],priority[maxn*5];int ch[maxn*5][2],cnt[maxn*5],size[maxn];Treap(){root=0;treapnode=1;priority[0]=INF;size[0]=0;}void up(int x){size[x]=size[ch[x][0]]+cnt[x]+size[ch[x][1]];}void rotate(int &x,int t){int y=ch[x][t];ch[x][t]=ch[y][t^1];ch[y][t^1]=x;up(x);up(y);x=y;}void insert(int &x,int k){if(x){if(key[x]==k)cnt[x]++;else {int t=key[x]<k;insert(ch[x][t],k);if(priority[ch[x][t]]<priority[x])rotate(x,t);}}else {x=treapnode++;key[x]=k;cnt[x]=1;priority[x]=rand();ch[x][0]=ch[x][1]=0;}up(x);}int getkth(int &x,int k){if(k<=size[ch[x][0]])return getkth(ch[x][0],k);k-=size[ch[x][0]]+cnt[x];if(k<=0)return key[x];return getkth(ch[x][1],k);}}T;int M,N,II,cc;int arr[33000];int u[33000];int main(){    while(~scanf("%d%d",&M,&N)){        II=0;cc=1;        for(int i=1;i<=M;i++) scanf("%d",arr+i);        for(int i=1;i<=N;i++)scanf("%d",u+i);        for(int i=1;i<=M;i++){           T.insert(T.root,arr[i]);           while(i==u[cc]){            cc++;II++;            printf("%d\n",T.getkth(T.root,II));          }     }  }    return 0;}