POJ-3784题解

题目描述

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

如果你题目看不懂的话，快试一试翻译

解题思路

我们使用维护两个堆一个大根堆一个小根堆的方法来进行动态中位数查找
当然我推荐直接使用STL的priority_queue，因为不容易写错^ ^
具体的维护过程如下：

我们现在读入了一个新的数$x$
　　1、如果他比大根堆的堆顶要小，也就是说明他比当前中位数要小，那么把$x$插入大根堆
　　2、如果他比大根堆的堆顶要大，也就是说明他比当前中位数要打，那么把$x$插入小根堆
　　3、在维护的过程中如果发现大根堆比小根堆少，弹出小根堆的堆顶插入大根堆直至符合要求
　　4、在维护的过程中如果发现大根堆的$size-$小根堆的$size>1$ ，弹出大根堆的堆顶假如小根堆直至符合要求

代码如下

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>

using namespace std;

priority_queue <int, vector<int>,greater<int> > q1;//这个是小根堆 分不清的时候可以跑个程序试一试
priority_queue <int> q2;

vector <int > a;

void Add(int x)
{
         
         
    if (q1.empty())
    {
         
         
        q1.push(x);
        return;
    }
    if (x>q1.top())
        q1.push(x);
    else
        q2.push(x);
    while (q2.size() > q1.size())
    {
         
         
        q1.push(q2.top());
        q2.pop();
    }
    while (q1.size() > q2.size() + 1)
    {
         
         
        q2.push(q1.top());
        q1.pop();
    }
}

int main()
{
         
         
    int t,k;
    cin>>t;
    while (t--)
    {
         
         
        while (q1.size()) q1.pop();
        while (q2.size()) q2.pop();
        a.clear();

        int  m,n;
        cin>>m>>n;
        for (int i=1;i<=n;i++)
        {
         
         
            cin>>k;
            Add(k);
            if (i%2!=0) a.push_back(q1.top());
        }
        cout<<m<<" "<<(n+1)/2<<endl;
        for (int i=0;i<a.size();i++)
        {
         
         
            if(i > 0 && i % 10 == 0) putchar('\n');
            if(i % 10) putchar(' ');
            printf("%d", a[i]);
        }
        cout<<endl;
    }
    return 0;
}