Running Median

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.

输入

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.

输出

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.

样例输入

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

样例输出

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

code

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
int main(){
	int t;
	scanf("%d", &t);
	while(t--){
		int id,n;
		scanf("%d%d", &id,&n);
		printf("%d %d
", id,(n+1)/2);
		priority_queue<int> l;
		priority_queue<int,vector<int>,greater<int>> r;
		for(int i = 1; i <= n; ++i){
			int x;
			scanf("%d", &x);
			if(l.size()>r.size()&&x<l.top())
				r.push(l.top()),l.pop(),l.push(x);
			else r.push(x);

			while(l.size()<r.size()) l.push(r.top()),r.pop();
			int j=(i+1)/2;
			if(i&1) printf("%d%s", l.top(),(i<n&&j%10!=0)?" ":"
");
		}
	}
	return 0;
}