이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define int long long
#define FOR(i, a, b) for (int i = a; i <= b; i++)
#define FORD(i, a, b) for (int i = b; i >= a; i --)
#define REP(i, n) for (int i = 0; i < n; ++i)
#define REPD(i, n) for (int i = n - 1; i >= 0; --i)
#define MASK(i) (1LL << (i))
#define BIT(x, i) (((x) >> (i)) & 1)
constexpr ll LINF = (1ll << 60);
constexpr int INF = (1ll << 30);
constexpr int Mod = 1e9 + 7;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
void setupIO(){
#define name "Whisper"
//Phu Trong from Nguyen Tat Thanh High School for gifted student
srand(time(NULL));
cin.tie(nullptr)->sync_with_stdio(false); cout.tie(nullptr);
//freopen(name".inp", "r", stdin);
//freopen(name".out", "w", stdout);
cout << fixed << setprecision(10);
}
template <class X, class Y>
bool minimize(X &x, const Y &y){
X eps = 1e-9;
if (x > y + eps) {x = y; return 1;}
return 0;
}
template <class X, class Y>
bool maximize(X &x, const Y &y){
X eps = 1e-9;
if (x + eps < y) {x = y; return 1;}
return 0;
}
int nArr, nGroup;
const int MAX = 1e5 + 5;
int a[MAX];
int dp[MAX][202];
int opt[MAX][202];
int s[MAX];
struct Line{
int a, b, id;
Line(): a(0), b(0), id(-1){}
Line(int _a, int _b, int _id): a(_a), b(_b), id(_id){}
int f(int x){ return a * x + b; }
double cross(Line o){
return (double) (b - o.b) / (double)(o.a - a);
}
};
//slope is increasing and query is also increasing
struct CHT{
deque<Line> q;
bool bad(Line a, Line b, Line c){
return a.cross(c) <= a.cross(b);
}
void addLine(Line line){
while (q.size() >= 2 && (bad(q.end()[-1], q.end()[-2], line) || (q.back().a == line.a && q.back().b <= line.b)))
q.pop_back();
q.push_back(line);
}
pair<int, int> query(int x){
while (q.size() >= 2 && q[0].f(x) <= q[1].f(x))
q.pop_front();
return make_pair(q[0].f(x), q[0].id);
}
};
void Whisper(){
cin >> nArr >> nGroup; ++nGroup;
for (int i = 1; i <= nArr; ++i) cin >> a[i], s[i] = s[i - 1] + a[i];
for (int i = 1; i <= nArr; ++i) dp[i][1] = s[i] * (s[nArr] - s[i]);
//s[nArr] * s[i] - s[nArr] * s[j] - s[i] * s[i] + s[i] * s[j]
//using s(j) as x we have equation
//s(j) * x + - s(nArr) * s(j) - s(i) * s(i) + s(nArr) * s(i) + dp(j, k - 1)
for (int k = 2; k <= nGroup; ++k){
CHT cht;
cht.addLine(Line(0, 0, -1));
for (int i = 1; i <= nArr; ++i){
pair<int, int> ret = cht.query(s[i]);
dp[i][k] = ret.first + s[nArr] * s[i] - s[i] * s[i];
opt[i][k] = ret.second;
cht.addLine(Line(s[i], - s[i] * s[nArr] + dp[i][k - 1], i));
}
}
cout << dp[nArr][nGroup] << '\n';
vector<int> ans;
int current = nArr;
while(nGroup > 0){
if (current != nArr)
ans.push_back(current);
current = opt[current][nGroup];
nGroup--;
}
reverse(ans.begin(), ans.end());
for (int&v : ans) cout << v << " ";
}
signed main(){
setupIO();
int Test = 1;
// cin >> Test;
for ( int i = 1 ; i <= Test ; i++ ){
Whisper();
if (i < Test) cout << '\n';
}
}
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