Absolute Minima（堆）

题目描述
There is a function f(x), which is initially a constant function f(x)=0.
We will ask you to process Q queries in order. There are two kinds of queries, update queries and evaluation queries, as follows:
An update query 1 a b: Given two integers a and b, let g(x)=f(x)+|x−a|+b and replace f(x) with g(x).
An evaluation query 2: Print x that minimizes f(x), and the minimum value of f(x). If there are multiple such values of x, choose the minimum such value.
We can show that the values to be output in an evaluation query are always integers, so we ask you to print those values as integers without decimal points.

Constraints
·All values in input are integers.
·1≤Q≤2×10⁵
·−10⁹≤a,b≤10⁹
·The first query is an update query.

输入
Input is given from Standard Input in the following format:
Q
Query1
:
QueryQ
See Sample Input 1 for an example.

输出
For each evaluation query, print a line containing the response, in the order in which the queries are given.

The response to each evaluation query should be the minimum value of x that minimizes f(x), and the minimum value of f(x), in this order, with space in between.

样例输入
4
1 4 2
2
1 1 -8
2

样例输出
4 2
1 -3

提示
In the first evaluation query, f(x)=|x−4|+2, which attains the minimum value of 2 at x=4.
In the second evaluation query, f(x)=|x−1|+|x−4|−6, which attains the minimum value of −3 when 1≤x≤4. Among the multiple values of x that minimize f(x), we ask you to print the minimum, that is, 1.

思路
本题可举例推出a序列的中位数即为f(x)的最小值,那么只需求得a序列的中位数，即可求解

代码实现

#pragma GCC optimize(3,"Ofast","inline")
#include
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int N=155;
const int M=600005;
const ll INF=1e14;
const ull sed=31;
const ll mod=1e9+7;
const double eps=1e-12;
typedef pair<int,int>P;

ll T,op,a,b,sum;
priority_queue<ll>q1;
priority_queue<ll,vector<ll>,greater<ll> >q2;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    cin>>T;
    while(T--)
    {
        cin>>op;
        if(op==1)
        {
            cin>>a>>b;
            q1.push(a);q2.push(a);
            sum+=b;
            if(q1.top()>q2.top())
            {
                sum+=abs(q2.top()-q1.top());
                a=q1.top();q1.pop();
                q1.push(q2.top());
                q2.pop();q2.push(a);
            }
        }
        else printf("%lld %lld\n",q1.top(),sum);
    }
    return 0;
}