POJ - 3784 Running Median 对顶堆 双优先级队列

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

#include<iostream>  
#include<cstdio>  
#include<algorithm>
#include<cmath>
#include<queue> 
using namespace std;  
#define ll long long  
const int len=1e5+4; 
int main()  
{  
   int t;
   cin>>t;
   while(t--)
   {
   		int c,n,u,v;
   		vector<int>ve;
   		scanf("%d%d",&c,&n);
   		priority_queue<int,vector<int>,greater<int> >q2;
   		priority_queue<int>q1;
   		for(int i=1;i<=n;++i)
   		{
   			scanf("%d",&u);
   			q2.push(u);
			if(i%2==1)
			{
				int x=q2.top();q2.pop();
				q1.push(x);
				x=q1.top();q1.pop();
				q2.push(x);
				x=q2.top();q2.pop();
				q1.push(x);x=q1.top();  
				ve.push_back(x);  
			} 
		}
		printf("%d %d\n",c,(n+1)/2);
		for(int i=0;i<ve.size();++i)
		{
			if((i+1)%10==0)printf("%d\n",ve[i]);
			else printf("%d ",ve[i]);
		}
		if(ve.size()%10)puts("");
   }
}