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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#112213#5466. Permutation Compression6arenRE 1ms3416kbC++178.2kb2023-06-10 17:45:012023-06-10 17:45:03

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2023-06-10 17:45:03]
  • 评测
  • 测评结果:RE
  • 用时:1ms
  • 内存:3416kb
  • [2023-06-10 17:45:01]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#ifdef LOCAL
#include <cp/debugger.hpp>
#else
#define debug(...) 42
#endif

#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()

typedef int64_t int64;
typedef pair<int, int> ii;

// source: https://github.com/ngthanhtrung23/ACM_Notebook_new/blob/master/DataStructure/LazySegTree.h
// Lazy Segment Tree, copied from AtCoder {{{
// Source: https://github.com/atcoder/ac-library/blob/master/atcoder/lazysegtree.hpp
// Doc: https://atcoder.github.io/ac-library/master/document_en/lazysegtree.html
//
// Notes:
// - Index of elements from 0
// - Range queries are [l, r-1]
// - composition(f, g) should return f(g())
//
// Tested:
// - https://oj.vnoi.info/problem/qmax2
// - https://oj.vnoi.info/problem/lites
// - (range set, add, mult, sum) https://oj.vnoi.info/problem/segtree_itmix
// - (range add (i-L)*A + B, sum) https://oj.vnoi.info/problem/segtree_itladder
// - https://atcoder.jp/contests/practice2/tasks/practice2_l
// - https://judge.yosupo.jp/problem/range_affine_range_sum

int ceil_pow2(int n) {
  int x = 0;
  while ((1U << x) < (unsigned int)(n)) x++;
  return x;
}
template <class S,                 // node data type
          S (*op)(S, S),           // combine 2 nodes
          S (*e)(),                // identity element
          class F,                 // lazy propagation tag
          S (*mapping)(F, S),      // apply tag F on a node
          F (*composition)(F, F),  // combine 2 tags
          F (*id)()                // identity tag
          >
struct LazySegTree {
  LazySegTree() : LazySegTree(0) {}
  explicit LazySegTree(int n) : LazySegTree(vector<S>(n, e())) {}
  explicit LazySegTree(const vector<S> &v) : _n((int)v.size()) {
    log = ceil_pow2(_n);
    size = 1 << log;
    d = std::vector<S>(2 * size, e());
    lz = std::vector<F>(size, id());
    for (int i = 0; i < _n; i++) d[size + i] = v[i];
    for (int i = size - 1; i >= 1; i--) {
      update(i);
    }
  }

  // 0 <= p < n
  void set(int p, S x) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    d[p] = x;
    for (int i = 1; i <= log; i++) update(p >> i);
  }

  // 0 <= p < n
  S get(int p) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    return d[p];
  }

  // Get product in range [l, r-1]
  // 0 <= l <= r <= n
  // For empty segment (l == r) -> return e()
  S prod(int l, int r) {
    assert(0 <= l && l <= r && r <= _n);
    if (l == r) return e();

    l += size;
    r += size;

    for (int i = log; i >= 1; i--) {
      if (((l >> i) << i) != l) push(l >> i);
      if (((r >> i) << i) != r) push((r - 1) >> i);
    }

    S sml = e(), smr = e();
    while (l < r) {
      if (l & 1) sml = op(sml, d[l++]);
      if (r & 1) smr = op(d[--r], smr);
      l >>= 1;
      r >>= 1;
    }

    return op(sml, smr);
  }

  S all_prod() { return d[1]; }

  // 0 <= p < n
  void apply(int p, F f) {
    assert(0 <= p && p < _n);
    p += size;
    for (int i = log; i >= 1; i--) push(p >> i);
    d[p] = mapping(f, d[p]);
    for (int i = 1; i <= log; i++) update(p >> i);
  }

  // Apply f on all elements in range [l, r-1]
  // 0 <= l <= r <= n
  void apply(int l, int r, F f) {
    assert(0 <= l && l <= r && r <= _n);
    if (l == r) return;

    l += size;
    r += size;

    for (int i = log; i >= 1; i--) {
      if (((l >> i) << i) != l) push(l >> i);
      if (((r >> i) << i) != r) push((r - 1) >> i);
    }

    {
      int l2 = l, r2 = r;
      while (l < r) {
        if (l & 1) all_apply(l++, f);
        if (r & 1) all_apply(--r, f);
        l >>= 1;
        r >>= 1;
      }
      l = l2;
      r = r2;
    }

    for (int i = 1; i <= log; i++) {
      if (((l >> i) << i) != l) update(l >> i);
      if (((r >> i) << i) != r) update((r - 1) >> i);
    }
  }

  // Binary search on SegTree to find largest r:
  //    f(op(a[l] .. a[r-1])) = true   (assuming empty array is always true)
  //    f(op(a[l] .. a[r])) = false    (assuming op(..., a[n]), which is out of bound, is always false)
  // template <bool (*g)(S)>
  // int max_right(int l) {
  //   return max_right(l, [](S x) { return g(x); });
  // }

  // Binary search on SegTree to find largest r:
  //    g(op(a[l] .. a[r-1])) = true   (assuming empty array is always true)
  //    g(op(a[l] .. a[r])) = false    (assuming op(..., a[n]), which is out of bound, is always false)
  template <class G>
  int max_right(int l, G g) {
    assert(0 <= l && l <= _n);
    assert(g(e()));
    if (l == _n) return _n;
    l += size;
    for (int i = log; i >= 1; i--) push(l >> i);
    S sm = e();
    do {
      while (l % 2 == 0) l >>= 1;
      if (!g(op(sm, d[l]))) {
        while (l < size) {
          push(l);
          l = (2 * l);
          if (g(op(sm, d[l]))) {
            sm = op(sm, d[l]);
            l++;
          }
        }
        return l - size;
      }
      sm = op(sm, d[l]);
      l++;
    } while ((l & -l) != l);
    return _n;
  }

  // Binary search on SegTree to find smallest l:
  //    f(op(a[l] .. a[r-1])) = true      (assuming empty array is always true)
  //    f(op(a[l-1] .. a[r-1])) = false   (assuming op(a[-1], ..), which is out of bound, is always false)
  template <class G>
  int min_left(int r, G g) {
    assert(0 <= r && r <= _n);
    assert(g(e()));
    if (r == 0) return 0;
    r += size;
    for (int i = log; i >= 1; i--) push((r - 1) >> i);
    S sm = e();
    do {
      r--;
      while (r > 1 && (r % 2)) r >>= 1;
      if (!g(op(d[r], sm))) {
        while (r < size) {
          push(r);
          r = (2 * r + 1);
          if (g(op(d[r], sm))) {
            sm = op(d[r], sm);
            r--;
          }
        }
        return r + 1 - size;
      }
      sm = op(d[r], sm);
    } while ((r & -r) != r);
    return 0;
  }

 private:
  int _n, size, log;
  vector<S> d;
  vector<F> lz;

  void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); }
  void all_apply(int k, F f) {
    d[k] = mapping(f, d[k]);
    if (k < size) lz[k] = composition(f, lz[k]);
  }
  void push(int k) {
    all_apply(2 * k, lz[k]);
    all_apply(2 * k + 1, lz[k]);
    lz[k] = id();
  }
};
// }}}

// Examples {{{
struct LazySegTreeOps {
  static const int NOT_ASSIGNED = -123123123;
  static int op(int lhs, int rhs) { return max(lhs, rhs); }
  static int e() { return INT_MIN; }
  static int mapping(int tag, int node) {
    if (tag == NOT_ASSIGNED) return node;
    return tag;
  }
  static int composition(int tag1, int tag2) { return tag1; }
  static int id() { return NOT_ASSIGNED; }
};
// LazySegTree<int, LazySegTreeOps::op, LazySegTreeOps::e, int, LazySegTreeOps::mapping, LazySegTreeOps::composition,
//             LazySegTreeOps::id>
//     seg_tree(a);
// }}}

void solve() {
  int n, m, k;
  cin >> n >> m >> k;
  vector<int> a(n), b(m);
  for (int &e : a) cin >> e;
  for (int &e : b) cin >> e;
  set<int> l;
  for (int i = 0; i < k; i++) {
    int x;
    cin >> x;
    l.insert(x);
  }
  LazySegTree<int, LazySegTreeOps::op, LazySegTreeOps::e, int, LazySegTreeOps::mapping, LazySegTreeOps::composition,
              LazySegTreeOps::id>
      seg_tree(a);
  vector<int> pos(n);
  for (int i = 0; i < n; i++) pos[a[i]] = i;
  for (int i = 0; i + 1 < m; i++) {
    if (pos[b[i]] > pos[b[i + 1]]) {
      cout << "NO\n";
      return;
    }
  }
  for (int e : b) {
    pos[e] = -1;
  }
  vector<int> removed;
  for (int e : a) {
    if (pos[e] != -1) removed.push_back(e);
  }
  sort(removed.rbegin(), removed.rend());
  for (int e : removed) {
    int right = seg_tree.max_right(pos[e], [&](int x) { return x <= e; });
    int left = seg_tree.min_left(pos[e] + 1, [&](int x) { return x <= e; }) - 1;
    int target = right - left - 1;
    auto it = l.upper_bound(target);
    if (it == l.begin()) {
      cout << "NO\n";
      return;
    }
    it--;
    l.erase(it);
    seg_tree.apply(pos[e], INT_MIN);
  }
  cout << "YES\n";
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3416kb

input:

3
5 2 3
5 1 3 2 4
5 2
1 2 4
5 5 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
3 2 2
3 1 2
3 2
2 3

output:

YES
YES
NO

result:

ok 3 lines

Test #2:

score: -100
Runtime Error

input:

100
2 1 2
2 1
2
1 1
2 1 2
1 2
1
2 2
2 1 1
1 2
1
2
6 1 5
3 4 2 5 6 1
3
5 2 1 1 1
6 1 6
2 1 3 6 4 5
1
4 1 2 2 1 4
3 3 2
2 1 3
2 1 3
2 2
1 1 1
1
1
1
1 1 1
1
1
1
2 1 2
2 1
2
1 2
4 4 3
2 1 3 4
2 1 3 4
4 3 1
1 1 1
1
1
1
6 5 1
6 2 5 4 3 1
6 2 4 3 1
4
1 1 1
1
1
1
6 5 3
3 6 1 4 5 2
3 6 1 4 2
3 3 4
4 3 4
3 4 ...

output:


result: