POJ 3784 Running Median【维护动态中位数】

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

传送门：http://poj.org/problem?id=3784

题意：每次读入一个整数序列，每当已经读入的整数个数为奇数时，输出已读入的整数构成的序列的中位数

思路：建立两个二叉堆：一个小根堆，一个大根堆。每次读入一个数X，若X比中位数小，则放入大根堆中，若X比中位数大，则放入小根堆中。如果某个时候，堆中的元素个数之差为2，则取出元素个数较多的那个堆的堆顶元素，放入另一个堆中，同时更新中位数。

代码：

 #include<bits/stdc++.h>

 using namespace std;

 int ans[];
 int main()

 {
     int T;
     scanf("%d", &T);
     while(T--)

     {

         int t;
         int n;
         int mid;

         scanf("%d%d%d", &t, &n, &mid);

         int cnt = ;
         ans[++cnt] = mid;

         priority_queue<int, vector<int>, greater<int> >s;//小根堆

         priority_queue<int, vector<int>, less<int> >b;//大根堆
         for(int i = ; i <= n; i++)
         {
             int temp;
             scanf("%d", &temp);

             if(temp > mid)
             {
                 s.push(temp);
                 if(s.size() - b.size() == )
                 {
                     b.push(mid);
                     mid = s.top();
                     s.pop();
                 }
             }
             else
             {
                 b.push(temp);
                 if(b.size() - s.size() == )
                 {
                     s.push(mid);
                     mid = b.top();
                     b.pop();
                 }
             }
             if(i %  )
                 ans[++cnt] = mid;
         }

         printf("%d %d\n", t, n /  +  );
         printf("%d", ans[] );

         for(int i = ; i <= cnt; i++)
         {
             printf(" %d", ans[i]);
             if(i %  == )
             {
                 printf("\n");
                 if(i != cnt)
                 {
                     printf("%d", ans[i + ]);
                     i++;
                 }
             }
         }
         printf("\n");
     }
 }