STL--H - Black Box（兩個優先隊列，求第k小的值）

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H - Black BoxTime Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64uSubmit 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 43 1 -4 2 8 -1000 21 2 6 6

Sample Output

3312

題意很麻煩：解釋一下資料 7 4 表示給出7個數，有4個詢問，下一行給出7個數，在下一行有4個詢問，“1”代表從頭到第一個元素中最小的值，“2”代表從頭到第二個元素中第二小的值，“6”代表從頭到第六個元素中第三小的值，“6”代表從頭到第六個元素中第四小的值，給出的詢問中 a[i] <= a[j] (i < j) ；

做法，定義兩個優先隊列，以大優先的p1，以小優先的p2，如果要求的是第x小的值，p1中存下(x-1)個小值，那麼第x個就是p2的隊首，在求第一個小的值，p1為空白，求完第一個小的值後，將p2的隊首放入p1，再來求第二小的值，讀取給出的數（a）時，如果a大於p1的隊首，那麼a放入p2，否則，將a放入p1，p1的隊首放入p2，保證p1的個數均比p2小，且為（x-1）個，讀取完數後p2的隊首就是第x小的數，輸出，再把p2的隊首放入p1，執行之前的操作，得到下一個要求的最小值。

用兩個優先隊列，分開整體的數組，得到第x小值

#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <algorithm>using namespace std;#define LL __int64priority_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;}