# AtCoder Regular Contest 074: D - 3N Numbers

,

## solution

$[0, r)$から$N$個選んだときの最大値と$[l, 3N)$から$N$個選んだときの最小値をそれぞれ$O(N \log N)$で計算しておいてまとめる。$O(N \log N)$。

$N$個選んだときの最大値最小値には、要素数が$N$という不変条件を持たせてpriority_queueとかを使う。

## implementation

#include <cstdio>
#include <vector>
#include <queue>
#define repeat(i,n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < int(n); ++(i))
#define repeat_reverse(i,n) for (int i = (n)-1; (i) >= 0; --(i))
using ll = long long;
using namespace std;
template <class T> inline void setmax(T & a, T const & b) { a = max(a, b); }
template <class T> using reversed_priority_queue = priority_queue<T, vector<T>, greater<T> >;

constexpr ll inf = ll(1e18)+9;
int main() {
// input
int n; scanf("%d", &n);
vector<ll> a(3*n); repeat (i,3*n) scanf("%lld", &a[i]);
// compute
vector<ll> front(3*n+1); {
reversed_priority_queue<ll> que;
repeat (i,3*n) {
que.push(a[i]);
ll b = 0;
if (que.size() == n+1) {
b = que.top(); que.pop();
}
front[i+1] = front[i] + a[i] - b;
}
}
vector<ll> back(3*n+1); {
priority_queue<ll> que;
repeat_reverse (i,3*n) {
que.push(a[i]);
ll b = 0;
if (que.size() == n+1) {
b = que.top(); que.pop();
}
back[i] = back[i+1] + a[i] - b;
}
}
ll result = - inf;
repeat_from (i,n,2*n+1) {
setmax(result, front[i] - back[i]);
}
// output
printf("%lld\n", result);
return 0;
}