POJ 1442-Black Box(優先隊列)

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Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 7436		Accepted: 3050

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

題意 ：給出n，m，然後接下一行輸入n個數，最後一行輸入m個數，要求對於最後一行的第i個數，輸出在原數列中前x個數中第i小的數，。好繞口，然後一開始用sort是各種TLE，好多大神用平衡樹過的，差點沒嚇哭我，然後，。。擼了兩個堆，一個最大堆，一個最小堆，注意維護好這兩個堆，其中最大堆放前k個最小的數，最小堆放剩餘的數，那麼最大堆的堆頂就是第k個最小的數。

#include <iostream>#include <cstring>#include <cstdio>#include <cctype>#include <cstdlib>#include <algorithm>#include <set>#include <vector>#include <string>#include <map>#include <queue>using namespace std;const int maxn= 30010;int a[maxn];priority_queue <int,vector<int>,less<int> > maxQ;priority_queue <int,vector<int>,greater<int> > minQ;int main(){int n,m,x;scanf("%d%d",&n,&m);for(int i=0;i<n;i++)        scanf("%d",&a[i]);int pos=0;for(int i=1;i<=m;i++){scanf("%d",&x);while(pos<x)minQ.push(a[pos++]);//新加元素放入最小堆while(maxQ.size()<i)//維護最大堆{maxQ.push(minQ.top());minQ.pop();}while(!minQ.empty()&&maxQ.top()>minQ.top())//維護兩個堆，嚴格保證maxQ.top()<=minQ.top(){int t1=maxQ.top(),t2=minQ.top();maxQ.pop();minQ.pop();maxQ.push(t2);minQ.push(t1);}printf("%d\n",maxQ.top());}return 0;}

POJ 1442-Black Box(優先隊列)