이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define sz(a) (int)a.size()
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define ff(i,a,b) for(int i=a;i<=b;i++)
#define fb(i,b,a) for(int i=b;i>=a;i--)
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,ll> pii;
const int maxn = 100005;
const ll inf = 1e14 + 5;
template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k) print the k-th smallest number in os(0-based)
int n, k;
int niz[maxn];
ll dp[maxn][105];
int main() {
ios::sync_with_stdio(false);
cout.tie(nullptr);
cin.tie(nullptr);
cin >> n >> k;
ff(i,1,n)cin >> niz[i];
ff(i,0,n) {
ff(j,0,k)dp[i][j] = inf;
}
int mx = 0;
ff(i,1,n){
mx = max(mx, niz[i]);
dp[i][1] = mx;
}
ff(j,2,k){
deque<pii> dq;
ff(i,1,n){
int x = niz[i];
ll mn = dp[i - 1][j - 1];
while(!dq.empty() && dq.back().fi <= x){
mn = min(mn, dq.back().se);
dq.pop_back();
}
if(dq.empty() || mn + x <= dq.back().fi + dq.back().se){
dq.push_back({x, mn});
}
if(i < j)continue;
dp[i][j] = dq.back().fi + dq.back().se;
}
}
cout << dp[n][k] << endl;
return 0;
}
/**
5 5
1 2 3 4 5
5 2
1 2 3 4 5
// probati bojenje sahovski ili slicno
**/
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