Submission #1061741

#TimeUsernameProblemLanguageResultExecution timeMemory
1061741VMaksimoski008K blocks (IZhO14_blocks)C++17
53 / 100
1055 ms40676 KiB
#include <bits/stdc++.h> 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, h; vector<ll> tree, lazy; SegTree(int _n) { n = _n; h = sizeof(int) * 8 - __builtin_clz(n); tree.resize(2*n+5); lazy.resize(n+5); } void apply(int p, ll v) { tree[p] += v; if(p < n) lazy[p] += v; } void build(int p) { while(p > 1) { p >>= 1; tree[p] = min(tree[p<<1], tree[p<<1|1]) + lazy[p]; } } void push(int p) { for(int s=h; s; --s) { int i = p >> s; if(lazy[i]) { apply(i<<1, lazy[i]); apply(i<<1|1, lazy[i]); lazy[i] = 0; } } } void update(int l, int r, ll v) { l+=n, r+=n; int l0=l, r0=r; for(; l<r; l>>=1, r>>=1) { if(l&1) apply(l++, v); if(r&1) apply(--r, v); } build(l0); build(r0-1); } ll query(int l, int r) { ll ans1 = 1e12, ans2 = 1e12; l+=n, r+=n; push(l); push(r-1); for(; l<r; l>>=1, r>>=1) { if(l&1) ans1 = min(ans1, tree[l++]); if(r&1) ans2 = min(ans2, tree[--r]); } return min(ans1, ans2); } void reset() { for(auto &x : tree) x = 0; for(auto &x : lazy) x = 0; } }; ll dp[maxn][105], v[maxn]; signed main() { ios_base::sync_with_stdio(false); cout.tie(0); cin.tie(0); int n, k; cin >> n >> k; for(int i=0; i<=n; i++) for(int j=0; j<=k; j++) dp[i][j] = 1e9; for(int i=1; i<=n; i++) cin >> v[i]; dp[0][0] = 0; ll mx = 0; for(int i=1; i<=n; i++) { mx = max(mx, v[i]); dp[i][1] = mx; } SegTree tree(n+1); for(int j=2; j<=k; j++) { vector<pii> st; st.push_back({ 1e9, 0 }); int SZ = 1; for(int i=1; i<=n; i++) { int last = st.back().second; while(SZ && st.back().first <= v[i]) { if(SZ > 1) tree.update(st[SZ-2].second+1, st[SZ-1].second+1, v[i]-st.back().first); st.pop_back(); SZ--; } st.push_back({ v[i], i }); SZ++; tree.update(last+1, st.back().second+1, v[i]); tree.update(i, i+1, dp[i-1][j-1]); dp[i][j] = tree.query(1, i+1); } tree.reset(); } cout << dp[n][k] << '\n'; return 0; }
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