本文主要是介绍poj 1442 Black Box，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：点击打开链接
Description
Our Black Box represents a primitive(原始的) database. It can save an integer(整数) array(数组) and has a special i variable(变量). At the initial moment Black Box is empty and i equals 0. This Black Box processes asequence(序列) of commands (transactions(交易)). There are two types of transactions:
ADD (x): put element(元素) 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(处于) 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
It is required to work out an efficient(有效率的) algorithm(算法) which treats a given sequence(序列) of transactions(交易). The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer(整数) arrays(数组):
1. A(1), A(2), ..., A(M): a sequence of elements(基础) which are being included into Black Box. A values are integers not exceeding(超过) 2 000 000 000 by their absolute(绝对的) 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(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.
ADD (x): put element(元素) 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(处于) 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(有效率的) algorithm(算法) which treats a given sequence(序列) of transactions(交易). The maximum number of ADD and GET transactions: 30000 of each type.
Let us describe the sequence of transactions by two integer(整数) arrays(数组):
1. A(1), A(2), ..., A(M): a sequence of elements(基础) which are being included into Black Box. A values are integers not exceeding(超过) 2 000 000 000 by their absolute(绝对的) 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(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(投入) 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(输出) 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
题意：第一行是序列 第二行的值i代表在输入第一行的第i个值时输出第
k大的数．k是从一开始的．（主要是看懂题意），用优先队列；
1＞用升序的优先队列存储，在将最小的放入降序的优先队列中，这样就能找到第k小的值了（降序队列中只方k个值）
<span style="font-size:24px;">#include <iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#include<cstdlib>
#include<algorithm>
using namespace std;
int a[30005];
int b[30005];
int main()
{int m,n;priority_queue< int ,vector<int >, less<int > >q;priority_queue< int ,vector<int >, greater<int > >p;cin>>m>>n;for(int i=1;i<=m;i++){cin>>a[i];}for(int j=1;j<=n;j++){cin>>b[j];}int i=0;int j,k;j=k=1;while(j<=n){if(i==b[j]){j++;if(q.size()<k){int t=p.top();q.push(t);p.pop();}int ans=q.top();cout<<ans<<endl;k++;}else{i++;if(q.size()<k){p.push(a[i]);int t=p.top();p.pop();q.push(t);}else if(q.top()>a[i]){int t=q.top();q.pop();p.push(t);q.push(a[i]);}else{p.push(a[i]);}}}return 0;
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