POJ1442——Black Box

Black Box Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 10621 Accepted: 4377

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：

给你一串数，有m(m<=30000)个数要成立一个序列，有n(n<=m)个询问，每次询问输入一个数，表示插入第几个数时输出第i个小的数。

以样例为例：输入第一个数  3  求起一个的数  为 3,  输入第二个数  1，求第二小的数为 3, 输入第6个数，因为6之前还有数没检测，所以要先进队，然后再输出第三小的数 1,。。。。。。。

例如S[1..n] = 1,2,3,4,5,6,7，index = 4，则大顶堆为{1,2,3}，小顶堆为{4,5,6,7}为什么要这样维护呢？因为当小堆最小的元素都大于大堆最大的元素时，那么序列中排第index个数就是小堆最小的数了。    我们假设第 k 次询问后，有以下情景：大顶堆：S[1..k]，堆顶元素为S[k]，小顶堆：S[k+1,n]，堆顶元素为S[k+1]，然后每当添加一个元素new时，先添加到大顶堆中，这时如果出现大顶堆数大于小顶堆的数时，理应交换。

#include <iostream> #include <cstdio> #include <queue> #include <vector> using namespace std; int main() { priority_queue<int, vector<int>,less<int> > mx; //大数优先 priority_queue<int, vector<int>,greater<int> >mn; //小数优先 int m, n, top = 0; cin>>m>>n; int *a = new int[m]; for( int i = 0;i < m;i++ ) cin>>a[i]; for( int j = 0;j < n;j++ ) { int x; cin>>x; int i; for( i = top;i < x;i++ ) { mn.push(a[i]); if( !mx.empty()&&mx.top()>mn.top() ) //核心 { int t1 = mn.top(); mn.pop(); int t2 = mx.top(); mx.pop(); mn.push(t2); mx.push(t1); } } top = i; cout<<mn.top()<<endl; mx.push(mn.top()); mn.pop(); //输出之后就没用了，出队 } return 0; }