poj 1442 Black Box

Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 8847 Accepted: 3636

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

331
2
题目大意：给出两种操作如下：( ADD  GET )    操作          当GET取第几大         所有已添加元素                                                           输出的元素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   用给出的输入实现以上操作。
原本是想队列写的，但超时:
#include<stdio.h>#include<queue>#include<iostream>#include<algorithm>using namespace std;int main(){int  n,m;while(scanf("%d%d",&n,&m)!=EOF){priority_queue<int ,vector<int >,greater<int > >q;priority_queue<int ,vector<int >,greater<int > >p;priority_queue<int ,vector<int >,less<int > >u;int a[40000],v,i;for(i=0;i<n;i++)scanf("%d",&a[i]); for(i=0;i<m;i++) { scanf("%d",&v); q.push(v);  } int temp,s,b;   int j=1; while(!q.empty() ) { int k=j; i=0; temp=q.top() ; q.pop() ; while(temp>i) {    p.push(a[i]);   i++; }while(!u.empty() )u.pop() ;while(k--){b=p.top();p.pop() ;u.push(b); }int s;             s=u.top();printf("%d\n",s);while(!p.empty() )p.pop() ; j++; } }} 
这是A的：（优先队列）
#include <stdio.h>#include <queue>using namespace std;int main(){    int a[30005],i,j,n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        int cut=0,x,c=0,t;        for(i=0; i<n; i++)        {            scanf("%d",&a[i]);        }        priority_queue <int , vector <int> , less<int> > p; //大顶堆        priority_queue <int , vector <int> , greater<int> >q;        for(i=0; i<m; i++)        {            scanf("%d",&x);            while(c<x)            {                q.push(a[c]);                c++;            }            while(!p.empty()&&p.top()>q.top())  //保证P的元素一定比q小            {                t=p.top();                p.pop();                p.push(q.top());                q.pop();                q.push(t);            }            printf("%d\n",q.top());            p.push(q.top());            q.pop();        }    }    return 0;}