結果
| 問題 | No.3446 Range Adjacent Differences |
| ユーザー |
|
| 提出日時 | 2026-02-18 22:00:50 |
| 言語 | C++23 (gcc 15.2.0 + boost 1.89.0) |
| 結果 |
TLE
|
| 実行時間 | - |
| コード長 | 3,405 bytes |
| 記録 | |
| コンパイル時間 | 7,545 ms |
| コンパイル使用メモリ | 392,504 KB |
| 実行使用メモリ | 16,200 KB |
| 最終ジャッジ日時 | 2026-02-18 22:01:03 |
| 合計ジャッジ時間 | 11,984 ms |
|
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 3 |
| other | TLE * 1 -- * 25 |
ソースコード
#include <bits/stdc++.h>
#include <atcoder/all>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using namespace atcoder;
using ll = long long;
using mint = modint998244353;
// using mint = modint1000000007;
template <typename T> using vec = vector<T>;
template <typename T> using pa = pair<T, T>;
#define REP(_1, _2, _3, _4, name, ...) name
#define REP1(i, n) for (auto i = decay_t<decltype(n)>{}; (i) < (n); ++(i))
#define REP2(i, l, r) for (auto i = (l); (i) < (r); ++(i))
#define REP3(i, l, r, d) for (auto i = (l); (i) < (r); i += (d))
#define rep(...) REP(__VA_ARGS__, REP3, REP2, REP1)(__VA_ARGS__)
#define rrep(i, r, l) for (int i = (r); i >= (l); --i)
template <typename T> bool chmax(T &a, const T &b) { return (a < b ? a = b, true : false); }
template <typename T> bool chmin(T &a, const T &b) { return (a > b ? a = b, true : false); }
constexpr int INF = 1 << 30;
constexpr ll LINF = 1LL << 60;
constexpr int mov[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
void solve();
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
// cin >> T;
while (T--) solve();
}
void solve() {
int N, Q;
cin >> N >> Q;
vec<int> A(N);
rep(i, N) cin >> A[i];
multiset<int> B, C;
auto add = [&] (int i) {
if (B.empty()) {
B.insert(A[i]);
return;
}
auto it = B.lower_bound(A[i]);
if (it == B.end()) {
C.insert(A[i] - *B.rbegin());
} else if (it == B.begin()) {
C.insert(*B.begin() - A[i]);
} else {
int r = *it, l = *--it;
C.erase(C.find(r - l));
C.insert(r - A[i]);
C.insert(A[i] - l);
}
B.insert(A[i]);
};
auto del = [&] (int i) {
if (B.size() == 1) {
B.erase(A[i]);
return;
}
B.erase(B.find(A[i]));
auto it = B.lower_bound(A[i]);
if (it == B.end()) {
C.erase(C.find(A[i] - *B.rbegin()));
} else if (it == B.begin()) {
C.erase(C.find(*B.begin() - A[i]));
} else {
int r = *it, l = *--it;
C.insert(r - l);
C.erase(C.find(r - A[i]));
C.erase(C.find(A[i] - l));
}
};
vec<int> L(Q), R(Q), X(Q), T(Q);
vec<int> ans(Q, -1);
rep(i, Q) {
int l, r, x;
char c;
cin >> l >> r >> x >> c;
l--;
L[i] = l;
R[i] = r;
X[i] = x;
T[i] = (c == 'L');
}
auto query = [&] (int i) {
if (T[i]) {
auto it = C.upper_bound(X[i]);
if (it != C.begin()) ans[i] = *--it;
} else {
auto it = C.lower_bound(X[i]);
if (it != C.end()) ans[i] = *it;
}
};
// Mo's
// 初めて自分で書くけど、割と楽~
// 高速化......?
int sq = max(1, (int)(N / sqrt(Q)));
vec<int> idx(Q);
ranges::iota(idx, 0);
ranges::sort(idx, [&] (int i, int j) {
if (L[i] / sq != L[j] / sq) return L[i] < L[j];
return R[i] < R[j];
});
int l = 0, r = 0;
for (auto i : idx) {
while (l > L[i]) add(--l);
while (r < R[i]) add(r++);
while (l < L[i]) del(l++);
while (r > R[i]) del(--r);
query(i);
}
rep(i, Q) cout << ans[i] << endl;
}