STL--H - Black Box（两个优先队列，求第k小的值）

H - Black Box

Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u

Submit Status

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

题意非常麻烦：解释一下数据 7 4 表示给出7个数，有4个询问，下一行给出7个数，在下一行有4个询问，“1”代表从头到第一个元素中最小的值。“2”代表从头到第二个元素中第二小的值。“6”代表从头到第六个元素中第三小的值，“6”代表从头到第六个元素中第四小的值。给出的询问中 a[i] <= a[j] (i < j) ；

做法，定义两个优先队列。以大优先的p1。以小优先的p2，假设要求的是第x小的值。p1中存下(x-1)个小值，那么第x个就是p2的队首，在求第一个小的值，p1为空，求完第一个小的值后，将p2的队首放入p1，再来求第二小的值。读取给出的数（a）时，假设a大于p1的队首，那么a放入p2。否则。将a放入p1，p1的队首放入p2，保证p1的个数均比p2小。且为（x-1）个，读取完数后p2的队首就是第x小的数，输出。再把p2的队首放入p1，运行之前的操作，得到下一个要求的最小值。

用两个优先队列。分开总体的数组，得到第x小值

#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;
#define LL __int64
priority_queue <LL> p1 ;
priority_queue <LL,vector<LL>,greater<LL> > p2 ;
LL a[6000000] ;
int main()
{
    int i , j , n , m , x ;
    LL temp ;
    while(scanf("%d %d", &n, &m)!=EOF)
    {
        while( !p1.empty() )
            p1.pop();
        while( !p2.empty() )
            p2.pop() ;
        for(i = 1 ; i <= n ; i++)
            scanf("%I64d", &a[i]);
        i = 1 ;
        while(m--)
        {
            scanf("%d", &x);
            for( ; i <= x ; i++)
            {
                if( p1.empty() || p1.top() < a[i] )
                    p2.push(a[i]);
                else
                {
                    p1.push(a[i]);
                    temp = p1.top() ;
                    p1.pop() ;
                    p2.push(temp);
                }
            }
            temp = p2.top();
            p2.pop() ;
            printf("%d\n", temp);
            p1.push(temp);
        }
    }
    return 0;
}