POJ 1442 Black Box（优先队列）

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 8710		Accepted: 3576

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

Source

Northeastern Europe 1996

      题意：第一行输入两个正整数n,m；n表示有n个数据，m代表m次操作，第二行输入n个数据，第三行有m个整数a1,a2,a3...am，每一个整数ai都表示输入第二行数据中的前ai个数据，然后求ai个数据中第h小的数字,h的值是第三行输入的整数的顺序（例如题面上第三行的数据中 “1的h的值为1””“2的h的值为2”“6的h的值为3”  “6（第二个6）的h的值为4”）

     思路：定义两个优先队列，一个从大到小排列q2，一个从小到大排列q1，将前h大的数放在q1中，剩下的放在q2中，每次插入优先队列都要对q2的第一个数据进行比较，如果比q2大，就直接插入到q1中，否则将q2的第一个数据插入到q1中，然后把比较的数放在q2中，当然q2的size小于h的时候直接存在q2中，总之这样做的目的是让序列中的前h大的数在q2中。

点击打开链接

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>

using namespace std;

int n,m;
int a[30011],b[30011];
int cnt;

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        priority_queue<int, vector<int>, greater<int> > q1;///从小到大排列
        priority_queue<int, vector<int>, less<int> > q2;///从大到小排列
        while(!q1.empty())
        {
            q1.pop();
        }
        while(!q2.empty())
        {
            q2.pop();
        }
        int ans = 0;
        for(int i=1; i<=m; i++)
        {
            scanf("%d",&cnt);
            for(int j=ans+1; j<=cnt; j++)
            {
                if(q2.size()<i)
                {
                    q2.push(a[j]);
                }
                else
                {
                    if(a[j]<q2.top())
                    {
                        q1.push(q2.top());
                        q2.pop();
                        q2.push(a[j]);
                    }
                    else
                    {
                        q1.push(a[j]);
                    }
                }
            }
            if(q2.size()<i)
            {
                int num = i - q2.size();
                for(int k=1; k<num; k++)
                {
                    b[k] = q1.top();
                    q1.pop();
                }
                printf("%d\n",q1.top());
                for(int k=1; k<num; k++)
                {
                    q1.push(b[k]);
                }
            }
            else
            {
                printf("%d\n",q2.top());
            }
            ans = cnt;
            if(!q1.empty())
            {
                q2.push(q1.top());
                q1.pop();
            }

        }
    }
    return 0;
}