思路分析:

    使用兩個堆, 一個最大堆儲存當前較小的一半, 一個最小堆儲存當前較大的一半, 每讀入一個數

動態維護這兩個堆, 當以讀入奇數個數時, 輸出當前最大堆堆頂即可. 具體如下AC程式碼所示:

//POJ3784_Running Median	
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <functional>
#include <queue>
using namespace std;
int main(){
	int P; scanf("%d", &P);
	while(P--){
		int sn, M; scanf("%d %d", &sn, &M); cout << sn << " " << M / 2 + 1 << endl;
		priority_queue<int, deque<int>, greater<int> > up; priority_queue<int> down;//始終維持down.size() >= up.size且up.size() >= down.size() - 1  
		for(int i = 1, t; i <= M; ++i){
			scanf("%d", &t);
			if(i & 1){
				if(up.empty()) down.push(t);
				else
					if(up.top() < t) down.push(up.top()), up.pop(), up.push(t);
					else down.push(t);
				cout << down.top() << " "; if(!((i + 1) % 20) || i == M) cout << endl;
			}else 
				if(down.top() > t) up.push(down.top()), down.pop(), down.push(t);
				else up.push(t);
		}
	}
	return 0;
}