poj1442 Black Box【优先队列，定义两个队列】

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 4
3 1 -4 2 8 -1000 2
1 2 6 6

3
3
1
2

#include <cstdio>
#include <queue>
using namespace std;
int main()
{
    int a[30005],i,j,n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
    	priority_queue <int , vector <int> , less<int> > p; //大的先
        priority_queue <int , vector <int> , greater<int> >q;//小的先
        int cut=0,x,c=0,t;
        for(i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        
        for(i=0; i<m; i++)
        {
            scanf("%d",&x);
            while(c<x)
            {
                q.push(a[c]);
                c++;
            }
            while(!p.empty()&&p.top()>q.top())  //保证第几小的数在q队列或p队列的顶部，然后计较一下两个的大小
            {
                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;
}