description: Monocarp plays a computer game (yet again!). This game has a unique trading mechanics.
  
  To trade with a character, Monocarp has to choose one of the items he possesses and trade it for some item the other character possesses. Each item has an integer price. If Monocarp's chosen item has price x, then he can trade it for any item (exactly one item) with price not greater than x+k.
  
  Monocarp initially has n items, the price of the i-th item he has is a_i. The character Monocarp is trading with has m items, the price of the i-th item they have is b_i. Monocarp can trade with this character as many times as he wants (possibly even zero times), each time exchanging one of his items with one of the other character's items according to the aforementioned constraints. Note that if Monocarp gets some item during an exchange, he can trade it for another item (since now the item belongs to him), and vice versa: if Monocarp trades one of his items for another item, he can get his item back by trading something for it.
  
  You have to answer q queries. Each query consists of one integer, which is the value of k, and asks you to calculate the maximum possible total cost of items Monocarp can have after some sequence of trades, assuming that he can trade an item of cost x for an item of cost not greater than x+k during each trade. Note that the queries are independent: the trades do not actually occur, Monocarp only wants to calculate the maximum total cost he can get.
  
  Input
  
  The first line contains three integers n, m and q (1 ≤ n, m, q ≤ 2 ⋅ 10^5).
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the prices of the items Monocarp has.
  
  The third line contains m integers b_1, b_2, ..., b_m (1 ≤ b_i ≤ 10^9) — the prices of the items the other character has.
  
  The fourth line contains q integers, where the i-th integer is the value of k for the i-th query (0 ≤ k ≤ 10^9).
  
  Output
  
  For each query, print one integer — the maximum possible total cost of items Monocarp can have after some sequence of trades, given the value of k from the query.
  
  Example
  
  Input
  
  
  3 4 5
  10 30 15
  12 31 14 18
  0 1 2 3 4
  
  
  Output
  
  
  55
  56
  60
  64
  64
solution: #include <bits/stdc++.h>
  using namespace std;
  const int N = 400 * 1000 + 13;
  int n, m, k;
  int a[N], b[N];
  int q[N];
  int p[N];
  multiset<int> wst[N], bst[N];
  long long sum;
  int getp(int a) { return a == p[a] ? a : p[a] = getp(p[a]); }
  void unite(int a, int b) {
    a = getp(a), b = getp(b);
    if (wst[a].size() + bst[a].size() < wst[b].size() + bst[b].size()) swap(a, b);
    for (auto it : wst[b]) wst[a].insert(it);
    for (auto it : bst[b]) bst[a].insert(it);
    wst[b].clear();
    bst[b].clear();
    while (!bst[a].empty() && !wst[a].empty() &&
           *bst[a].begin() < *wst[a].rbegin()) {
      sum -= *bst[a].begin();
      sum += *wst[a].rbegin();
      bst[a].insert(*wst[a].rbegin());
      wst[a].insert(*bst[a].begin());
      bst[a].erase(bst[a].begin());
      wst[a].erase(--wst[a].end());
    }
    p[b] = a;
  }
  long long ans[N];
  struct event {
    int x, t, i;
  };
  int main() {
    scanf("%d%d%d", &n, &m, &k);
    for (int i = 0; i < int(n); i++) scanf("%d", &a[i]);
    for (int i = 0; i < int(m); i++) scanf("%d", &b[i]);
    for (int i = 0; i < int(k); i++) scanf("%d", &q[i]);
    vector<pair<int, int>> tot;
    for (int i = 0; i < int(n); i++) tot.push_back({a[i], 1});
    for (int i = 0; i < int(m); i++) tot.push_back({b[i], 0});
    sort(tot.begin(), tot.end());
    sum = accumulate(a, a + n, 0ll);
    for (int i = 0; i < int(n + m); i++) {
      p[i] = i;
      wst[i].clear();
      bst[i].clear();
      if (tot[i].second)
        bst[i].insert(tot[i].first);
      else
        wst[i].insert(tot[i].first);
    }
    vector<event> ev;
    for (int i = 0; i < int(n + m - 1); i++)
      ev.push_back({tot[i + 1].first - tot[i].first, 0, i});
    for (int i = 0; i < int(k); i++) ev.push_back({q[i], 1, i});
    sort(ev.begin(), ev.end(), [](const event &a, const event &b) {
      if (a.x != b.x) return a.x < b.x;
      return a.t < b.t;
    });
    for (auto it : ev) {
      if (it.t == 0)
        unite(it.i, it.i + 1);
      else
        ans[it.i] = sum;
    }
    for (int i = 0; i < int(k); i++) printf("%lld\n", ans[i]);
  }
  