数据结构（堆）：POJ 1442 Black Box

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10658		Accepted: 4390

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
　　水题瞬秒……

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 #include <queue>
 using namespace std;
 const int maxn=;
 int a[maxn],t[maxn],n,m;
 priority_queue<int>A;
 priority_queue<int,vector<int>,greater<int> >B;
 int main(){
     scanf("%d%d",&n,&m);
     for(int i=;i<=n;i++)scanf("%d",&a[i]);
     for(int i=;i<=m;i++)scanf("%d",&t[i]);
     for(int i=,p=;i<=m;i++){
         while(p!=t[i])B.push(a[++p]);
         while(A.size()<1ul*i){
             A.push(B.top());
             B.pop();
         }
         while(B.size()&&A.top()>B.top()){
             A.push(B.top());
             B.push(A.top());
             A.pop();B.pop();
         }
         printf("%d\n",A.top());
     }
     return ;
 }