luogu 3466 对顶堆

显然答案是将一段区间全部转化成了其中位数
这样的话，需要维护一个数据结构支持查询当前所有数中位数
对顶堆
用两个堆
将 < 中位数的数放入大根堆
将 > 中位数的数放入小根堆
这样就会存在删除操作
删除的时候
由于无法快速删除
只需做个标记，标记该数被删除了一次
并且堆的实际大小也应该另外记录维护
在标记时需要更改相应的堆的大小与权值
答案就非常显然了

#include <bits/stdc++.h>

using namespace std;

#define gc getchar()
inline int read() {
    int x = ; char c = gc;
    while(c < '' || c > '') c = gc;
    while(c >= '' && c <= '') x = x *  + c - '', c = gc;
    return x;
}

#define Rep(i, a, b) for(int i = a; i <= b; i ++)
#define Fin(str) freopen(str, "r", stdin)
#define Fout(str) freopen(str, "w", stdout)
#define E return
#define LL long long

const int N = 1e5 + ;

int n, k;
int A[N], use[N * ];

priority_queue <int, vector<int>, less<int> > Big;
priority_queue <int, vector<int>, greater<int> > Small;

int Size1, Size2;
LL Sum1, Sum2;

void Update() {
    while(Big.size() && use[Big.top()]) use[Big.top()] --, Big.pop();
    while(Small.size() && use[Small.top()]) use[Small.top()] --, Small.pop();
}

void Out();

void Add(int x) {
    Update();
    while(Size1 < (k + ) /  && Small.size()) {
        int num = Small.top();
        Small.pop();
        Size1 ++, Sum1 += num;
        Size2 --, Sum2 -= num;
        Big.push(num);
        Update();
    }
    Update();
    if(Size1 < (k + ) / ) {
        Size1 ++, Sum1 += x, Big.push(x); return ;
    }
    Update();
    int num = Big.top();
    if(x < num) {
        Small.push(num);
        Big.pop();
        Size2 ++, Sum2 += num;
        Sum1 += (x - num);
        Big.push(x);
    } else {
        Size2 ++, Sum2 += x, Small.push(x);
    }
    Update();
}

void Del(int x) {
    use[x] ++;
    int top1 = Big.top();
    if(x <= top1) Size1 --, Sum1 -= x;
    else Size2 --, Sum2 -= x;
}

LL Calc() {
    Update();
    LL mid = Big.top();
    return 1ll * mid * Size1 - 1ll * Sum1 + 1ll * Sum2 - 1ll * mid * Size2;
}

int main() {
    n = read(), k = read();
    Rep(i, , n) A[i] = read();
    Rep(i, , k) Add(A[i]);
    LL Answer = Calc();
    int bef = ;
    int flag = k, mid = Big.top();
    Rep(i, k + , n) {
        Del(A[++ bef]);
        Add(A[i]);
        LL Ans = Calc();
        if(Ans < Answer) {
            flag = i, mid = Big.top();
            Answer = Ans;
        }
    }
    cout << Answer << "\n";
    Rep(i, , n) {
        if(i <= flag && i >= flag - k + ) {
            cout << mid << "\n";
        } else cout << A[i] << "\n";
    }
    return ;
}