제출 #1173169

#제출 시각아이디문제언어결과실행 시간메모리
1173169AmirAli_H1Growing Vegetables is Fun 5 (JOI24_vegetables5)C++20
9 / 100
185 ms66348 KiB
// In the name of Allah #include <bits/stdc++.h> using namespace std; typedef long long int ll; typedef long double ld; typedef pair<int, int> pii; typedef pair<ll, ll> pll; typedef complex<ld> cld; #define all(x) (x).begin(),(x).end() #define len(x) ((ll) (x).size()) #define F first #define S second #define pb push_back #define sep ' ' #define endl '\n' #define Mp make_pair #define kill(x) cout << x << '\n', exit(0) #define set_dec(x) cout << fixed << setprecision(x); #define file_io(x,y) freopen(x, "r", stdin); freopen(y, "w", stdout); mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); const int maxn = (1 << 20) + 7; const int oo = 1e9 + 4; struct node { int lazy; pii min, max; }; int n, m; pll A[maxn]; ll A1[maxn], A2[maxn]; int arr[maxn]; pii val1[maxn], val2[maxn]; node t[2 * maxn], null, f; int cnt, M[maxn]; vector<ll> lsx; bool cmp(int i, int j) { return (A[i] < A[j]); } node fx(node a, node b, int lazy) { node res; res.lazy = 0; res.min = min(a.min, b.min); res.max = max(a.max, b.max); res.lazy += lazy; res.min.F += lazy; res.max.F += lazy; return res; } void build(int v, int tl, int tr, bool R) { t[v].lazy = 0; if (tr - tl == 1) { int j = tl; if (R == 1) j = (j + m); if (j >= n) j -= n; if (!M[j]) t[v] = null; else { t[v].min = Mp(0, tl); t[v].max = Mp(0, tl); } return ; } int mid = (tl + tr) / 2; build(2 * v + 1, tl, mid, R); build(2 * v + 2, mid, tr, R); t[v] = fx(t[2 * v + 1], t[2 * v + 2], t[v].lazy); } void del_val(int v, int tl, int tr, int i) { if (i >= tr || i < tl) return ; if (tr - tl == 1) { t[v] = null; return ; } int mid = (tl + tr) / 2; if (i < mid) del_val(2 * v + 1, tl, mid, i); else del_val(2 * v + 2, mid, tr, i); t[v] = fx(t[2 * v + 1], t[2 * v + 2], t[v].lazy); } void add_val(int v, int tl, int tr, int l, int r, int x) { l = max(l, tl); r = min(r, tr); if (l >= tr || r <= tl || l >= r) return ; if (l == tl && r == tr) { t[v].lazy += x; t[v].min.F += x; t[v].max.F += x; return ; } int mid = (tl + tr) / 2; add_val(2 * v + 1, tl, mid, l, r, x); add_val(2 * v + 2, mid, tr, l, r, x); t[v] = fx(t[2 * v + 1], t[2 * v + 2], t[v].lazy); } node get_val(int v, int tl, int tr, int l, int r) { l = max(l, tl); r = min(r, tr); if (l >= tr || r <= tl || l >= r) return null; if (l == tl && r == tr) return t[v]; int mid = (tl + tr) / 2; return fx(get_val(2 * v + 1, tl, mid, l, r), get_val(2 * v + 2, mid, tr, l, r), t[v].lazy); } void addx(int l, int r) { add_val(0, 0, n, l, r, 1); add_val(0, 0, n, l + n, n, 1); } node getx(int l, int r) { return fx(get_val(0, 0, n, l, r), get_val(0, 0, n, l + n, n), 0); } void delx(int j, int R) { del_val(0, 0, n, j); if (R == 1) j = (j + m); if (j >= n) j -= n; cnt -= M[j]; M[j] = 0; } bool check(int x) { for (int i = 0; i < n; i++) { int j1 = lower_bound(A1, A1 + m, A[i].F - x) - A1; int j2 = upper_bound(A1, A1 + m, A[i].F + x) - A1 - 1; val1[i] = Mp(j1, j2); j1 = lower_bound(A2, A2 + m, A[i].F - x) - A2; j2 = upper_bound(A2, A2 + m, A[i].F + x) - A2 - 1; val2[i] = Mp(j1, j2); if (val1[i].F > val1[i].S && val2[i].F > val2[i].S) return 0; } cnt = n; fill(M, M + n, 1); if (n <= 2000) { build(0, 0, n, 0); for (int j = 0; j < n; j++) { int i = arr[j]; f = getx(i - m + 1, i + 1); while (f.min.F < val1[i].F) { int j = f.min.S; delx(j, 0); if (cnt <= 0) return 0; f = getx(i - m + 1, i + 1); } while (f.max.F > val1[i].S) { int j = f.max.S; delx(j, 0); if (cnt <= 0) return 0; f = getx(i - m + 1, i + 1); } addx(i - m + 1, i + 1); } build(0, 0, n, 1); for (int j = 0; j < n; j++) { int i = arr[j]; f = getx(i - m + 1, i + 1); while (f.min.F < val2[i].F) { int j = f.min.S; delx(j, 1); if (cnt <= 0) return 0; f = getx(i - m + 1, i + 1); } while (f.max.F > val2[i].S) { int j = f.max.S; delx(j, 1); if (cnt <= 0) return 0; f = getx(i - m + 1, i + 1); } addx(i - m + 1, i + 1); } } return 1; } int cal(int i) { ll mx = 0; lsx.clear(); for (int j = i; j < i + n; j++) { int R = j; if (R >= 2 * n) R -= 2 * n; lsx.pb(A[R].F); } sort(all(lsx)); for (int i = 0; i < n; i++) mx = max(mx, abs(A1[i] - lsx[i])); lsx.clear(); for (int j = i + n; j < i + (2 * n); j++) { int R = j; if (R >= 2 * n) R -= 2 * n; lsx.pb(A[R].F); } sort(all(lsx)); for (int j = 0; j < n; j++) mx = max(mx, abs(A2[j] - lsx[j])); return mx; } void solve() { cin >> m; n = 2 * m; for (int i = 0; i < n; i++) { cin >> A[i].F; A[i].S = (2 * min(i, n - i)) + (i / m); } for (int i = 0; i < m; i++) cin >> A1[i]; for (int i = 0; i < m; i++) cin >> A2[i]; iota(arr, arr + n, 0); sort(arr, arr + n, cmp); sort(A1, A1 + m); sort(A2, A2 + m); int l = -1, r = cal(rng() % n); while (r - l > 1) { int mid = (l + r) / 2; if (check(mid)) r = mid; else l = mid; } cout << r << endl; } int main() { ios::sync_with_stdio(false); cin.tie(0); cout.tie(0); null.min = Mp(oo, -1); null.max = Mp(-oo, -1); int T = 1; while (T--) { solve(); } return 0; }
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