Running Median 动态求中位数
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
用两个堆， 大顶堆和小顶堆
每次输入一个数，如果这个数比当前的中位数大，就存入小顶堆中， 否则就存入大顶堆。
然后调整， 小顶堆元素的个数要等于大顶堆的元素个数，或者比其多1。
如果小顶堆的元素太多，就塞到大顶堆里，反之亦然
这样一来就会发现。小顶堆的元素比所有大顶堆的元素都大， 而且小顶堆的堆顶就是中位数。
/*Running Median 动态求中位数
这个用到了堆，
用两个堆， 大顶堆和小顶堆
每次输入一个数，如果这个数比当前的中位数大，就存入小顶堆中， 否则就存入大顶堆。
这两个队列，一个从小到大排序，另一个从大到小
*/
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
using namespace std;
/*
priority_queue<int>q;//大根堆
priority_queue<int,vector<int>,greater<int> >Q;//小根堆
*/
priority_queue<int,vector<int>,greater<int>> q1; //从小到大排列
//q1放的是第一个数（当作中位数处理）和比其大的数
priority_queue<int,vector<int>,less<int>> q2; //从大到小排列
vector<int> g;
void add(int x)
{
if(q1.empty())
{
q1.push(x);
return; //保证第一个数可以被记录下来
}
if(x>q1.top)
q1.push(x);
else
q2.push(x);
while(q1.size()<q2.size())
{
q1.push(q2.top());
q2.pop();
}
while(q1.size()>q2.size())
{
q2.push(q1.top());
q1.pop();
}
}
int main()
{
int t,cas,n,x;
cin>>t;
while(t--)
{
//开始保证这两个队列为空
while(!q1.empty()) q1.pop();
while(!q2.empty()) q2.pop();
g.clear();
cin>>cas>>n;
for(int i=0;i<n;i++)
{
cin>>x;
add(x);
if(i%2==0)
g.push_back(q1.top());
}
cout<<cas<<(n+1)/2<<endl;
for(int i = 0; i < g.size(); i++)
{
if(i > 0 && i % 10 == 0) putchar('
');
if(i % 10) putchar(' ');
cout<<g[i];
}
cout<<endl;
}
return 0;
}