HDU - 3282 Running Median （動態中位數+優先隊列）

題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=3282點擊打開鏈接

Running Median

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1199    Accepted Submission(s): 431

Problem 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

 

Source

2009 Greater New York Regional

每到奇數時求中位數

用大根堆和小根堆儲存

維護小根堆數量奇數時爲大根堆+1偶數時相同 

每次遇到奇數查詢小根堆頂元素即可

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        priority_queue<int > qmax;
        priority_queue<int ,vector<int > ,greater <int > >qmin;
        int cnt;
        int n;
        scanf("%d%d",&cnt,&n);
        printf("%d %d\n",cnt,(n/2+1));
        for(int i=1;i<=n;i++)
        {
            int mid;
            scanf("%d",&mid);
            if(qmin.size()==0)
                qmin.push(mid);
            else
            {
                if(mid>=qmin.top())
                    qmin.push(mid);
                else
                    qmax.push(mid);
            }
            while(qmin.size()!=qmax.size()+1&&qmin.size()!=qmax.size())
            {
                if(qmin.size()>qmax.size())
                {
                    int mid=qmin.top();
                    qmin.pop();
                    qmax.push(mid);
                }
                else
                {
                    int mid=qmax.top();
                    qmax.pop();
                    qmin.push(mid);
                }
            }
            if(i&1)
                cout << qmin.top();
            if((i+1)%20==0||i==n)
                cout << endl;
            else if(i&1)
                cout << " ";
        }
    }
}