POJ Black Box 1442(优先队列)

                                                                                              Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 10302		Accepted: 4237

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(下降).

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) aninequality(不平等) 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

网上大神说的http://blog.youkuaiyun.com/lyhvoyage/article/details/18360655

题意：给定M个数，每次可以插入序列一个数；再给N个数，表示在插入第几个数时输出一个数，第一次输出序列中最小的，第二次输出序列中第二小的……以此类推，直到输出N个数。

分析：因为输出时是按照先输出最小的，再输出第二小这样的方式输出的，相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的，而是不断的在更新，所以用数组是无法实现的。我们可以用优先队列来做。

定义两个优先队列，一个用来存储前k小的数，大数在前，小数在后；另一个优先队列第k+1小到最大的数，小数在前，大数在后。每次拿到一个数，先判断第一个优先队列中的数满不满k个，如果不满k个，则直接把这个数压入到第一个队列；如果满k个，判断这个数和第一个优先队列中的第一个数的大小：如果比第一个数大，就压入第二个优先队列；如果比第一个数小，就把第一个优先队列的队首元素弹出压入第二个队列，把这个新数压入第一个优先队列。

输出时，如果第一个优先队列里的元素个数小于k，则先把第二个优先队列里的队首元素弹出压入第一个优先队列，然后输出第一个优先队列的队首元素；如果满k个，则直接输出第一个优先队列的队首元素。

我也是按照他说的写


#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <queue>
#include <algorithm>
using namespace std;
int main()
{
    int a[30010],b[30010];
    int n,m,i,j,k,x;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    for(i=0;i<m;i++)
    {
        scanf("%d",&b[i]);
    }
    priority_queue<int,vector<int>,greater<int> >q1;//k+1...da
    priority_queue<int,vector<int>,less<int> >q2;//da
    i=0;j=0;k=1;
    while(j<m)
    {
        if(i==b[j])
        {
            j++;
            if(q2.size()<k)
            {
                x=q1.top();
                q1.pop();
                q2.push(x);
            }
            x=q2.top();
            printf("%d\n",x);
            k++;
        }
        else
        {
            if(q2.size()<k)
            {
                q1.push(a[i]);
                x=q1.top();
                q1.pop();
                q2.push(x);
            }
            else if(q2.top()>a[i])
            {
                x=q2.top();
                q2.pop();
                q1.push(x);
                q2.push(a[i]);
            }
            else
            {
                q1.push(a[i]);
            }
            i++;
        }
    }
    return 0;
}