This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#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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