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ID题目提交者结果用时内存语言文件大小提交时间测评时间
#423403#7108. Couleurucup-team3215RE 1ms5500kbC++204.9kb2024-05-27 23:46:242024-05-27 23:46:25

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你现在查看的是最新测评结果

  • [2024-05-27 23:46:25]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:5500kb
  • [2024-05-27 23:46:24]
  • 提交

answer

#include <algorithm>
#include <array>
#include <iostream>
#include <map>
#include <random>
#include <set>
#include <vector>

using namespace std;

constexpr int N = 1e5;

int ft[N];

void upd(int i, int x, int n) {
  for (; i < n; i |= i + 1) ft[i] += x;
}

int que(int i) {
  int x{};
  for (; i; i &= i - 1) x += ft[i - 1];
  return x;
}

struct Node;

struct Ptr {
  int i{-1};

  Node* operator->();
  operator bool() { return ~i; };
};

struct Node {
  int x, y, sz;
  Ptr l, r;
};

Node nodes[N];

Node* Ptr::operator->() { return nodes + i; }

int sget(Ptr p) { return p? p->sz: 0; }
Ptr upd(Ptr p) { return p->sz = sget(p->l) + sget(p->r) + 1, p; }

Ptr merge(Ptr l, Ptr r) {
  if (!l || !r) return l ?: r;
  if (l->y < r->y) return l->r = merge(l->r, r), upd(l);
  else             return r->l = merge(l, r->l), upd(r);
}

Ptr extract(Ptr p, array<int, 2> rb, vector<array<int, 2>>& keys, vector<Ptr>& extracted, int64_t& invs, int ord) {
  if (!p || keys.empty() || keys.back() >= rb) return p;
  p->l = extract(p->l, array{p->x, p.i}, keys, extracted, invs, ord);
  while (p && keys.size() && array{p->x, p.i} == keys.back()) {
    invs += ord + sget(p->l);
    extracted.push_back(p);
    p = merge(p->l, p->r);
    keys.pop_back();
    p->l = extract(p->l, array{p->x, p.i}, keys, extracted, invs, ord);
  }
  return p? p->r = extract(p->r, rb, keys, extracted, invs, ord + sget(p->l) + 1), upd(p): Ptr{};
}

Ptr build(vector<Ptr>& sorted) {
  if (sorted.empty()) return {};
  vector<Ptr> st{sorted[0]};
  sorted[0]->l = {};
  for (int i = 1; i < sorted.size(); ++i) {
    auto p = sorted[i];
    Ptr left;
    while (st.size() && st.back()->y > p->y) {
      st.back()->r = left;
      left = upd(st.back());
      st.pop_back();
    }
    p->l = left;
    st.push_back(p);
  }
  st.back()->r = {};
  while (st.size() > 1) {
    auto r = st.back();
    st.pop_back();
    st.back()->r = upd(r);
  }
  sorted.clear();
  return upd(st[0]);
}

int main() {
  cin.tie(0)->sync_with_stdio(0);
  if (mt19937 rng(23123213); 1)
  for (auto& x: nodes) x.y = rng();
  for (int tc = (cin >> tc, tc); tc--; ) {
    int n; cin >> n;
    for (int i = 0; i < n; ++i) cin >> nodes[i].x, nodes[i].sz = 1, ft[i] = 0;
    map<int, pair<Ptr, int64_t>> byleft;
    multiset<int64_t> costs;
    vector<Ptr> tobuild(n);
    vector<array<int, 2>> keys;
    {
      vector<array<int, 2>> sorted(n);
      for (int i = 0; i < n; ++i) sorted[i] = {nodes[i].x, i};
      sort(sorted.begin(), sorted.end());
      for (int i = 0; i < n; ++i) tobuild[i] = {sorted[i][1]};
      int64_t cost = 0;
      for (int i = n; i--; ) upd(sorted[i][1], 1, n), cost += que(sorted[i][1]);
      fill_n(ft, n, 0);
      byleft[0] = {build(tobuild), cost};
      costs.insert(cost);
    }
    for(int i = 0; i < n; ++i) {
      int64_t mx = *costs.rbegin(), q;
      cout << mx << ' ';
      cin >> q; q ^= mx; --q;
      auto nit = byleft.upper_bound(q), it = prev(nit);
      int l = it->first, r = nit == byleft.end()? n: nit->first;
      auto [p, c] = it->second;
      byleft.erase(it);
      costs.erase(costs.find(c));
      if (q - l < r - q) {
        int64_t linvs = 0, invs = 0, invs2 = 0;
        for (int i = l; i < q; ++i) keys.push_back({nodes[i].x, i});
        sort(keys.rbegin(), keys.rend());
        for (int i = 0; i < keys.size(); ++i) upd(keys[i][1] - l, 1, q - l), linvs += que(keys[i][1] - l);
        fill_n(ft, q - l, 0);
        if (l != q) {
          p = extract(p, array<int, 2>{N + 1}, keys, tobuild, invs, 0);
          byleft[l] = {build(tobuild), linvs};
          costs.insert(linvs);
        }
        keys.push_back({nodes[q].x, q});
        p = extract(p, array<int, 2>{N + 1}, keys, tobuild, invs2, 0);
        tobuild.clear();
        byleft[q] = {};
        if (q + 1 != r) {
          byleft[q + 1] = {p, c - linvs - invs - invs2};
          costs.insert(c - linvs - invs - invs2);
        }
      } else {
        int64_t rinvs = 0, invs = 0, invs2 = 0;
        for (int i = q + 1; i < r; ++i) keys.push_back({nodes[i].x, i});
        sort(keys.rbegin(), keys.rend());
        for (int i = 0; i < keys.size(); ++i) upd(keys[i][1] - q - 1, 1, r - q - 1), rinvs += que(keys[i][1] - q - 1);
        fill_n(ft, r - q - 1, 0);
        if (q + 1 != r) {
          p = extract(p, array<int, 2>{N + 1}, keys, tobuild, invs, 0);
          byleft[q + 1] = {build(tobuild), rinvs};
          costs.insert(rinvs);
        }
        keys.push_back({nodes[q].x, q});
        p = extract(p, array<int, 2>{N + 1}, keys, tobuild, invs2, 0);
        tobuild.clear();
        byleft[q] = {};
        if (l != q) {
          byleft[l] = {p, c - rinvs - ((q + 1 - l) * int64_t(r - q - 1) - invs) - (q - l - invs2)};
          costs.insert(c - rinvs - ((q + 1 - l) * int64_t(r - q - 1) - invs) - (q - l - invs2));
        }
      }
    }
    cout << '\n';
  }
}

详细

Test #1:

score: 100
Accepted
time: 1ms
memory: 5500kb

input:

3
5
4 3 1 1 1
5 4 5 3 1
10
9 7 1 4 7 8 5 7 4 8
21 8 15 5 9 2 4 5 10 6
15
4 8 8 1 12 1 10 14 7 14 2 9 13 10 3
37 19 23 15 7 2 10 15 2 13 4 5 8 7 10

output:

7 0 0 0 0 
20 11 7 2 0 0 0 0 0 0 
42 31 21 14 14 4 1 1 1 0 0 0 0 0 0 

result:

ok 3 lines

Test #2:

score: -100
Runtime Error

input:

11116
10
10 5 10 3 6 4 8 5 9 8
31 27 24 11 12 3 0 2 3 1
10
8 2 7 2 8 10 1 10 9 10
6 5 2 13 2 1 0 1 3 1
10
7 10 7 6 1 3 10 6 7 9
21 18 10 1 6 5 4 8 9 10
10
2 10 4 8 8 5 7 2 6 7
20 10 9 1 15 0 4 2 9 7
10
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 10
10
1 2 3 4 5 6 7 8 9 10
6 3 5 2 7 10 9 1 4 8
10
1 10 1 3...

output:


result: