Black Box

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 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.

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

7 43 1 -4 2 8 -1000 21 2 6 6

3312

题意：用测试数据解释。给出7个数，4次操作。第一次操作是1,说明前一个数中第一小的数是多少，第二次操作是2,说明前2个数中第2小的是多少，
第三次操作是6,说明前6个数中第3小的是多少。第四次操作是6,说明前6个中第4小的是多少。

思路：用到两个堆，一个最大堆，一个最小堆。其中最大堆用来维护，最小堆用来求k小数。 

 1 /*====================================================================== 2  *           Author :   kevin 3  *         Filename :   BlackBox.cpp 4  *       Creat time :   2014-07-28 15:46 5  *      Description : 6  ========================================================================*/ 7 #include <iostream> 8 #include <algorithm> 9 #include <cstdio>10 #include <cstring>11 #include <queue>12 #include <cmath>13 #define clr(a,b) memset(a,b,sizeof(a))14 #define M 3000515 using namespace std;16 int s[M];17 struct cmp18 {19     bool operator() (const int& x,const int& y){20         return x > y;21     }22 };23 int main(int argc,char *argv[])24 {25     int n,m;26     while(scanf("%d%d",&n,&m)!=EOF){27         priority_queue<int>max;28         priority_queue<int,vector<int>,cmp>min;29         for(int i = 0; i < n; i++){30             scanf("%d",&s[i]);31         }32         int a,t = 0,minheap,maxheap;33         for(int i = 0; i < m; i++){34             scanf("%d",&a);35             for(int i = t; i < a; i++){36                 min.push(s[i]);37                 if(!max.empty()){38                     minheap = min.top();39                     maxheap = max.top();40                     if(minheap < maxheap){41                         min.pop(); max.pop();42                         min.push(maxheap);43                         max.push(minheap);44                     }45                 }46             }47             t = a;48             printf("%d\n",min.top());49             max.push(min.top());50             min.pop();51         }52     }53     return 0;54 }