POJ-1442 Black Box(手写堆优化)

Posted 2022-09-16 MichaelZona

Black Box

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11150		Accepted: 4554

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​​

 

心累到现在都没过……

分析：因为输出时是按照先输出最小的，再输出第二小这样的方式输出的，相当于依次输出一个有序序列中的值。但因为这个序列不是固定不变的，而是不断的在更新，所以用数组是无法实现的。我们可以用优先队列来做。

定义两个优先队列，一个用来存储前k小的数，大数在前，小数在后；另一个优先队列第k+1小到最大的数，小数在前，大数在后。每次拿到一个数，先判断第一个优先队列中的数满不满k个，如果不满k个，则直接把这个数压入到第一个队列；如果满k个，判断这个数和第一个优先队列中的第一个数的大小：如果比第一个数大，就压入第二个优先队列；如果比第一个数小，就把第一个优先队列的队首元素弹出压入第二个队列，把这个新数压入第一个优先队列。

输出时，如果第一个优先队列里的元素个数小于k，则先把第二个优先队列里的队首元素弹出压入第一个优先队列，然后输出第一个优先队列的队首元素；如果满k个，则直接输出第一个优先队列的队首元素。

1 #include "bits/stdc++.h"
 2 #define mem(a,b) memset(a,b,sizeof(a))
 3 using namespace std;
 4 typedef long long LL;
 5 const int MAX=30005;
 6 int n,m;
 7 int a[MAX],r[MAX];
 8 struct Que
 9     int h[MAX*10];
10     int n;
11     void ini()mem(h,0),n=0;
12     void heapify1(int x)
13         int child=x*2,key=h[x];
14         while (child<=n)
15             if (child<n && h[child]<h[child+1]) child++;
16             if (key<h[child])h[x]=h[child];x=child;child=x*2;
17             else break;
18         
19         h[x]=key;
20     
21     void insert1(int key)
22         int x=++n;
23         while (x>1)
24             if (key>h[x/2]) h[x]=h[x/2],x/=2;
25             else break;
26         
27         h[x]=key;
28     
29     void del1()
30         if (n==1) n=0;
31         else h[1]=h[n--],heapify1(1);
32     
33     void heapify2(int x)
34         int child=x*2,key=h[x];
35         while (child<=n)
36             if (child<n && h[child]>h[child+1]) child++;
37             if (h[x]>h[child])h[x]=h[child];x=child;child=x*2;
38             else break;
39         
40         h[x]=key;
41     
42     void insert2(int key)
43         int x=++n;
44         while (x>1)
45             if (key<h[x/2]) h[x]=h[x/2],x/=2;
46             else break;
47         
48         h[x]=key;
49     
50     void del2()
51         if (n==1) n=0;
52         else h[1]=h[n--],heapify2(1);
53     
54 q1,q2;
55 int main()
56     freopen ("box.in","r",stdin);
57     freopen ("box.out","w",stdout);
58     int i,j,k;
59     while (~scanf("%d%d",&n,&m))
60         q1.ini();q2.ini();
61         for (i=1;i<=n;i++)
62          scanf("%d",a+i);
63         for (j=1;j<=m;j++)
64          scanf("%d",r+j);
65         sort(r+1,r+m+1);
66         int index(1);
67         for (i=1;i<=n;i++)
68             if (index>n)
69              break;
70             if (q1.n<index)
71                 q1.insert1(a[i]);
72             else if (a[i]<q1.h[1])
73                 int zt=q1.h[1];
74                 q1.del1();
75                 q1.insert1(a[i]);
76                 q2.insert2(zt);
77             
78             else 
79              q2.insert2(a[i]);
80             while (i==r[index])
81                 printf("%d\\n",q1.h[1]);
82                 if (q2.n>0)
83                     q1.insert1(q2.h[1]);
84                     q2.del2();
85                 
86                 index++;
87             
88         
89     
90     return 0;
91