제출 #1067553

#제출 시각아이디문제언어결과실행 시간메모리
1067553VMaksimoski008Financial Report (JOI21_financial)C++17
45 / 100
4069 ms318240 KiB
#include <bits/stdc++.h> //#define int long long using namespace std; using ll = long long; using pii = pair<int, int>; using pll = pair<ll, ll>; const int mod = 1e9 + 7; const int LOG = 20; const int maxn = 1e5 + 5; struct SegTree { int n; vector<int> tree; SegTree(int _n) : n(_n), tree(4*_n+5) {} void update(int u, int tl, int tr, int p, int v) { if(tl == tr) { tree[u] = max(tree[u], v); } else { int tm = (tl + tr) / 2; if(p <= tm) update(2*u, tl, tm, p, v); else update(2*u+1, tm+1, tr, p, v); tree[u] = max(tree[2*u], tree[2*u+1]); } } int query(int u, int tl, int tr, int l, int r) { if(l > tr || tl > r) return 0; if(l <= tl && tr <= r) return tree[u]; int tm = (tl + tr) / 2; return max(query(2*u, tl, tm, l, r), query(2*u+1, tm+1, tr, l, r)); } void update(int p, int v) { update(1, 0, n-1, p, v); } int query(int l, int r) { return query(1, 0, n-1, l, r); } }; int dp[7005][7005], pref[7005][7005], suf[7005][7005]; signed main() { int n, d; cin >> n >> d; vector<int> v(n+1); set<int> s; for(int i=1; i<=n; i++) cin >> v[i], s.insert(v[i]); vector<int> comp(s.begin(), s.end()); for(int i=1; i<=n; i++) v[i] = lower_bound(comp.begin(), comp.end(), v[i]) - comp.begin(); int m = s.size(); //zemame bilo koja podsekvenca if(d == n) { int ans = 0; SegTree tree(n); for(int i=n; i>=1; i--) { ans = max(ans, tree.query(v[i]+1, n-1) + 1); tree.update(v[i], tree.query(v[i]+1, n-1) + 1); } cout << ans << '\n'; return 0; } //ako pocneme od i pa odime do n, odogovorot e kolku pati max ke se smeni //stack??? //od desno na levo cuvame >= stack if(d == 1) { int ans = 0; stack<pii> st; for(int i=n; i>=1; i--) { while(!st.empty() && st.top().first <= v[i]) st.pop(); st.push({ v[i], i }); ans = max(ans, (int)st.size()); } cout << ans << '\n'; return 0; } for(int i=1; i<=n; i++) { dp[i][v[i]] = 1; for(int k=i-1; k>=max(1, i-d); k--) for(int j=v[i]; j<m; j++) dp[i][j] = max(dp[i][j], dp[k][j]); for(int k=i-1; k>=max(1, i-d); k--) dp[i][v[i]] = max(dp[i][v[i]], (v[i] >= 1 ? pref[k][v[i]-1] + 1 : 1)); pref[i][0] = dp[i][0]; for(int j=1; j<m; j++) pref[i][j] = max(pref[i][j-1], dp[i][j]); } int ans = 0; for(int i=0; i<m; i++) ans = max(ans, dp[n][i]); cout << ans << '\n'; return 0; }
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...
#Verdict Execution timeMemoryGrader output
Fetching results...