Submission #57413

#TimeUsernameProblemLanguageResultExecution timeMemory
57413BenqK blocks (IZhO14_blocks)C++14
100 / 100
352 ms50968 KiB
#include <bits/stdc++.h> #include <ext/pb_ds/tree_policy.hpp> #include <ext/pb_ds/assoc_container.hpp> using namespace std; using namespace __gnu_pbds; typedef long long ll; typedef long double ld; typedef complex<ld> cd; typedef pair<int, int> pi; typedef pair<ll,ll> pl; typedef pair<ld,ld> pd; typedef vector<int> vi; typedef vector<ld> vd; typedef vector<ll> vl; typedef vector<pi> vpi; typedef vector<pl> vpl; typedef vector<cd> vcd; template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>; #define FOR(i, a, b) for (int i=a; i<(b); i++) #define F0R(i, a) for (int i=0; i<(a); i++) #define FORd(i,a,b) for (int i = (b)-1; i >= a; i--) #define F0Rd(i,a) for (int i = (a)-1; i >= 0; i--) #define sz(x) (int)(x).size() #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound #define all(x) x.begin(), x.end() const int MOD = 1000000007; const ll INF = 1e18; const int MX = 100001; int N,K, dp[101][MX],A[MX]; vpi bes; void ins(int x, int y) { if (x == 0 && y > 0) return; while (sz(bes) && bes.back().f >= dp[x][y]) bes.pop_back(); bes.pb({dp[x][y],-MOD}); } void process(int x) { int mn = MOD; while (sz(bes) && bes.back().s <= x) { mn = bes.back().f; bes.pop_back(); } if (!sz(bes) || bes.back().f+bes.back().s > mn+x) bes.pb({mn,x}); } void solve(int x) { bes.clear(); FOR(i,x,N+1) { ins(x-1,i-1); process(A[i]); dp[x][i] = bes.back().f+bes.back().s; } } int main() { ios_base::sync_with_stdio(0); cin.tie(0); cin >> N >> K; FOR(i,1,N+1) cin >> A[i]; FOR(i,1,K+1) solve(i); cout << dp[K][N]; } /* Look for: * the exact constraints (multiple sets are too slow for n=10^6 :( ) * special cases (n=1?) * overflow (ll vs int?) * array bounds */
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