POJ 3784 Running Median（动态维护中位数）

Posted 2021-01-12 wkfvawl

Description

For this problem, you will write a program that reads in a sequence of 32-bit signed integers. After each odd-indexed value is read, output the median (middle value) of the elements received so far.

Input

The first line of input contains a single integer P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by an odd decimal integer M, (1 ≤ M ≤ 9999), giving the total number of signed integers to be processed. The remaining line(s) in the dataset consists of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than 10 values.

Output

For each data set the first line of output contains the data set number, a single space and the number of medians output (which should be one-half the number of input values plus one). The output medians will be on the following lines, 10 per line separated by a single space. The last line may have less than 10 elements, but at least 1 element. There should be no blank lines in the output.

Sample Input

3 
1 9 
1 2 3 4 5 6 7 8 9 
2 9 
9 8 7 6 5 4 3 2 1 
3 23 
23 41 13 22 -3 24 -31 -11 -8 -7 
3 5 103 211 -311 -45 -67 -73 -81 -99 
-33 24 56

Sample Output

1 5
1 2 3 4 5
2 5
9 8 7 6 5
3 12
23 23 22 22 13 3 5 5 3 -3 
-7 -3

题目意思：对于n个数，每输入到奇数个数，便求一次中位数。

解题思路：我开始就直接暴力，到了奇数位，sort一下，找出中位数，这样在UVAlive上没有超时，但在poj上是超时的，看了看网上的代码才知道处理这样一个动态的中位数使用堆来维护，而这个堆使用优先队列来模拟，一周内第二次见到优先队列了。维护一个大根堆和一个小根堆，且保证大根堆里的所有数都比小根堆里的所有数小，而且大根堆的大小等于小根堆或者大1，则大根堆堆顶就是中位数。对于新插入的数，与中位数比较决定插入哪个堆中，插入之后维护一下两个堆的大小。每次维护需要进行常数个堆上的操作，所以复杂度O(nlogn)。

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<queue>
 4 using namespace std;
 5 priority_queue<int,vector<int>,greater<int> > bh;///从小到大
 6 priority_queue<int,vector<int>,less<int> > sh;///从大到小
 7 int main()
 8 {
 9     int t,n,num;
10     int i,j,m,p,q,x,y,z,K;
11     scanf("%d",&t);
12     while (t--)
13     {
14         scanf("%d%d",&num,&n);
15         printf("%d %d
",num,n/2+1);
16         while (!bh.empty())
17         {
18             bh.pop();
19         }
20         while (!sh.empty())
21         {
22             sh.pop();
23         }
24         for (i=1; i<=n; i++)
25         {
26             scanf("%d",&x);
27             if (sh.empty()||sh.top()>x)
28             {
29                 sh.push(x);///放入小堆
30             }
31             else
32             {
33                 bh.push(x);///放入大堆
34             }
35             if (sh.size()>(i+1)/2)
36             {
37                 x=sh.top();
38                 sh.pop();
39                 bh.push(x);
40             }
41             if (sh.size()<(i+1)/2)
42             {
43                 x=bh.top();
44                 bh.pop();
45                 sh.push(x);
46             }
47             if (i%2)
48             {
49                 printf("%d",sh.top());
50                 if (i==n||(i+1)%20==0)
51                 {
52                     printf("
");
53                 }
54                 else
55                 {
56                     printf(" ");
57                 }
58             }
59         }
60     }
61     return 0;
62 }