제출 #566447

#제출 시각아이디문제언어결과실행 시간메모리
566447dxz05Global Warming (CEOI18_glo)C++14
27 / 100
717 ms35912 KiB
//#pragma GCC optimize("Ofast,O2,O3,unroll-loops") //#pragma GCC target("avx2") #include <bits/stdc++.h> using namespace std; void debug_out() { cerr << endl; } template<typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cerr << "[" << H << "]"; debug_out(T...); } #ifdef dddxxz #define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__) #else #define debug(...) 42 #endif #define SZ(s) ((int)s.size()) #define all(x) (x).begin(), (x).end() #define lla(x) (x).rbegin(), (x).rend() #define bpc(x) __builtin_popcount(x) #define bpcll(x) __builtin_popcountll(x) clock_t startTime; double getCurrentTime() { return (double) (clock() - startTime) / CLOCKS_PER_SEC; } #define MP make_pair typedef long long ll; mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count()); const double eps = 0.000001; const int MOD = 1e9 + 7; const int INF = 1000000101; const long long LLINF = 1223372000000000555; const int N = 1e6 + 3e2; const int M = 1248; struct SegTree{ vector<int> tree; int size = 1; SegTree(int n){ size = n; tree.assign(4 * n + 5, 0); } void update(int v, int tl, int tr, int pos, int val){ if (tl == tr){ tree[v] = max(tree[v], val); return; } int tm = (tl + tr) >> 1; if (pos <= tm) update(v << 1, tl, tm, pos, val); else update(v << 1 | 1, tm + 1, tr, pos, val); tree[v] = max(tree[v << 1], tree[v << 1 | 1]); } void update(int pos, int val){ update(1, 1, size, pos, val); } int get(int v, int tl, int tr, int l, int r){ if (l <= tl && tr <= r) return tree[v]; if (tl > r || tr < l) return 0; int tm = (tl + tr) >> 1; return max(get(v << 1, tl, tm, l, r), get(v << 1 | 1, tm + 1, tr, l, r)); } int get(int l, int r){ return get(1, 1, size, l, r); } }; void solve(int TC) { int n, lim; cin >> n >> lim; vector<int> a(n); for (int i = 0; i < n; i++){ cin >> a[i]; } map<int, int> mp; for (int i : a) mp[i] = 0, mp[i + lim] = 0; int foo = 0; for (auto &now : mp) now.second = ++foo; SegTree st1(2 * n), st2(2 * n); int ans = 0; vector<int> dp1(n, 0), dp2(n, 0); for (int i = 0; i < n; i++){ int x = mp[a[i]]; dp1[i] = st1.get(1, x - 1) + 1; st1.update(x, dp1[i]); int y = mp[a[i] + lim]; int cur = st1.get(1, y - 1) + 1; st1.update(y, cur); ans = max(ans, cur); } cout << ans << endl; return; for (int i = n - 1; i >= 0; i--){ int x = mp[a[i]]; dp2[i] = st2.get(x + 1, n) + 1; st2.update(x, dp2[i]); } for (int i = 0; i < n; i++){ } } int main() { startTime = clock(); ios_base::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr); #ifdef dddxxz freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout); #endif int TC = 1; //cin >> TC; for (int test = 1; test <= TC; test++) { //debug(test); //cout << "Case #" << test << ": "; solve(test); } #ifdef dddxxz cerr << endl << "Time: " << int(getCurrentTime() * 1000) << " ms" << endl; #endif return 0; }
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