优先队列模拟最大堆和最小堆，poj 1442 Black Box

Black Box

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 5619 Accepted: 2265

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

Sample Output

3312

Source

Northeastern Europe 1996

 

题目链接：http://poj.org/problem?id=1442

 

 

 

/*题意：按顺序输入一些数（add），对应以后的每次get操作，第i次get x的意思是在输入的前x个数里找到第i小的数分析：数据很复杂，要求对每次get进行高效的回应，选择用堆来做，建立一个最大堆和一个最小堆，最大堆里放前i-1小的数，最小堆里放剩下的；  于是要求的第i小的数就是最小堆的堆顶元素。注意：动态维护。每次先把数据放进最小堆里，然后动态维护最小堆的堆顶要比最大堆的堆顶还要大。然后看是否达到get的数值，是，则输出，然后转移给最大堆元素达到i-1个。*/#include<cstdio>#include<queue>#include<iostream>#include<functional>//用仿函数greater<> 要用这个头文件#define max 30005using namespace std;priority_queue<long, vector<long>, greater<long> > min_heap;//优先队列里用仿函数，成最小堆。priority_queue<long> max_heap;long add[max],Get[max];int main(){    int M,N;    while(~scanf("%d %d",&M,&N)){    for(int i=1;i<=M;i++) scanf("%ld",&add[i]);    for(int i=1;i<=N;i++) scanf("%ld",&Get[i]);    int flag=1;    Get[0]=0;    for(int i=1;i<=M;i++){        min_heap.push(add[i]);//先放入最小堆        if(!max_heap.empty()){            if(max_heap.top()>min_heap.top())/*最大堆的堆顶比最小堆的堆顶大时，把最小堆的堆顶放到最大堆里，再把最大堆新的堆顶放到最小堆里，保证与最初的定义不矛盾*/{                max_heap.push(min_heap.top());                min_heap.pop();                min_heap.push(max_heap.top());                max_heap.pop();            }        }        while(Get[flag]==(min_heap.size()+max_heap.size()))//达到get的时间时输出最小堆的堆顶{            printf("%ld\n",min_heap.top());            flag++;//flag表示题目中的i的含义            while(max_heap.size()<flag-1)//保证最大堆里有flag-1个元素。{                max_heap.push(min_heap.top());                min_heap.pop();            }        }    }}    return 0;}