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>
#define ll long long
#define fi first
#define se second
#define endl '\n'
#define MASK(i) (1LL << (i))
#define ull unsigned long long
#define ld long double
#define pb push_back
#define all(x) (x).begin() , (x).end()
#define BIT(x , i) ((x >> (i)) & 1)
#define TASK "task"
#define sz(s) (int) (s).size()
using namespace std;
const int mxN = 1e5 + 227;
const int inf = 1e9 + 277;
const int mod = 1e9 + 7;
const ll infll = 1e18 + 7;
const int base = 4000;
const int LOG = 50;
const int block = 700;
template <typename T1, typename T2> bool minimize(T1 &a, T2 b) {
if (a > b) {a = b; return true;} return false;
}
template <typename T1, typename T2> bool maximize(T1 &a, T2 b) {
if (a < b) {a = b; return true;} return false;
}
inline namespace _LineContainer {
bool _Line_Comp_State;
struct Line {
// k is slope, m is intercept, p is intersection point
mutable ll k, m, p, idx;
bool operator<(const Line& o) const {
return _Line_Comp_State ? p < o.p : k < o.k;
}
};
struct LineContainer : multiset<Line> {
long long div(long long a, long long b) { return a / b - ((a ^ b) < 0 && a % b); }
bool isect(iterator x, iterator y) {
if (y == end()) {
x->p = LLONG_MAX;
return false;
}
if (x->k == y->k)
x->p = x->m > y->m ? LLONG_MAX : -LLONG_MAX;
else
x->p = div(y->m - x->m, x->k - y->k);
return x->p >= y->p;
}
void add(long long k, long long m, int idx) {
auto z = insert({k, m, 0, idx}), y = z++, x = y;
while (isect(y, z))
z = erase(z);
if (x != begin() && isect(--x, y))
isect(x, y = erase(y));
while ((y = x) != begin() && (--x)->p >= y->p)
isect(x, erase(y));
}
pair<ll, int> query(long long x) {
assert(!empty());
_Line_Comp_State = 1;
auto l = *lower_bound({0,0,x,0});
_Line_Comp_State = 0;
return make_pair(l.k * x + l.m, l.idx);
}
};
}
int n;
int k;
int a[mxN];
int pre[mxN];
int trace[207][mxN];
ll dp[2][mxN];
LineContainer lc;
vector<int> ans;
void solve()
{
cin >> n >> k;
for(int i = 1; i <= n; i++) cin >> a[i];
for(int i = 1; i <= n; i++) pre[i] = pre[i - 1] + a[i];
memset(dp, -0x3f, sizeof(dp));
ll oo = dp[0][0];
for(int i = 1; i <= n; i++) dp[0][i] = 0;
for(int i = 1; i <= k; i++) {
lc.clear();
for(int j = 0; j <= n; j++) dp[i & 1][j] = oo;
for(int j = 1; j <= n; j++) {
lc.add(pre[j - 1], dp[(i - 1) & 1][j - 1] - 1LL * pre[j - 1] * pre[j - 1], j - 1);
pair<ll, int> tmp = lc.query(pre[j]);
if(maximize(dp[i & 1][j], tmp.fi) == true) {
trace[i][j] = tmp.se;
}
}
}
cout << dp[k & 1][n] << endl;
int x = k;
int y = n;
while(x > 0) {
int tmp = trace[x][y];
if(y != n) ans.pb(y);
--x;
y = tmp;
}
ans.pb(y);
reverse(all(ans));
for(auto it : ans) cout << it << ' ';
}
// dp[i][k] = dp[j][k - 1] + (pre[i] - pre[j]) * pre[j]
// dp[i][k] = dp[j][k - 1] - pre[j] ^ 2 + pre[i] * pre[j]
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
//freopen("task.inp", "r", stdin);
//freopen("task.out", "w", stdout);
int tc = 1;
//cin >> tc;
while(tc--) {
solve();
}
return 0;
}
//3 5 7
//4 1 3 4 0 2 3
//4 1 3 | 4 0 2 3
//4 1 3 | 4 0 | 2 3
//4 1 3 | 4 0 | 2 3
//
//
//3 * 14 = 42
//5 * 12 = 60
// 8 * 4 + 12 * 5
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|
Fetching results... |