Algorithm学习笔记 --- Black Box-复制linux系统-程序博客网

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 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.

Write to the output Black Box answers sequence for a given sequence of transactions, one number each line.

Northeastern Europe 1996

题意：

此题的意思是一个模拟的数据库黑盒子，里面可以进行数据操作（还有一些都是废话就不解释了），

M（要插入多少个数） N（当队列里面的数满足输入的时候第几次GET就输出第几个最小的数）

解题分析：

此题在刚看的时候我也被题里面的这么多英文吓住了，后来在网上找了找翻译，并且自己把给的那几组数据从重新在纸上理解了一下，发现明白了。大家来看：

1 ADD(3)      0 3//在没有元素的情况下输入一个3   2 GET         1 3                                    3 //此时里面有了一个元素是3，要弹出第一个小的元素，先观察里面的元素是否够多，不够多继续添加，一下以此类推，在这里可以看出里面的元素刚好一个正好弹出，并且弹出最小的一个元素就是3.3 ADD(1)      1 1, 3   //接下来有又添加一个元素1，里面的元素是1，34 GET         2 1, 3                                 3 //现在里面有两个元素，再次观察这里的元素是否够两个，这里正好是两个，并且弹出第二小的元素是3.5 ADD(-4)     2 -4, 1, 3   //接下来加入-4  里面是-4，1，3，在这里观察这里的元素是否够6个，若不够继续添加。若够就弹出第三次的GET（第几次GET就输出第几小的数）元素6 ADD(2)      2 -4, 1, 2, 3   //接下来加入2，里面是-4，1，2，3，元素还是不够7 ADD(8)      2 -4, 1, 2, 3, 8   //接下来添加8.里面一共5个元素，-4，1，2，3，8还是不够继续添加。8 ADD(-1000)  2 -1000, -4, 1, 2, 3, 8//接下来添加-1000，此时里面刚好6个，下一步就不用继续添加，   9 GET         3 -1000, -4, 1, 2, 3, 8                1//此时元素刚好是6个，在这里是第三个GET,那就弹出第三个数（此处已经排序） 10 GET       4 -1000, -4, 1, 2, 3, 8                2 //再此也是输出第四个数11 ADD(2)     4 -1000, -4, 1, 2, 2, 3, 8

结束。

对于这列可以用单调队列(堆处理的数据可以用C++中的ＳＴＬ模板)，比较简单。

代码：

#include<iostream>
#include<cstdio>
#include<queue> //队列的头文件
#include<functional>//greater<>的头文件
#define MAXN 30005
using namespace std;
priority_queue<int, vector<int>, greater<int> > min_heap;//优先队列按照由小到大顺序；
priority_queue<int> max_heap; //priority_queue调用 STL里面的 make_heap(), pop_heap(), push_heap()；
int Add[MAXN],Get[MAXN]; //创建两个队列；
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; //设计一个标志位
Get[0]=0; //初始化输出队列
for(int i=1;i<=M;i++)
{
min_heap.push(Add[i]); //先放入Add队列
if(!max_heap.empty()) //判断是不是空的
{
if(max_heap.top()>min_heap.top()) //最大堆的堆顶比最小堆的堆顶大时
{
max_heap.push(min_heap.top()); //吧对小队列的头弹出放到最大队列里进行比较
min_heap.pop(); //把最小队列的头弹出
min_heap.push(max_heap.top()); //把最大队列的头部弹出放到最小队列里进行比较
max_heap.pop(); //把最大队列的头弹出
}
}
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("%d\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;
}