Submission #447992

#	Time	Username	Problem	Language	Result	Execution time	Memory
447992		gcmango	Split the sequence (APIO14_sequence)	C++17	89 / 100	1337 ms	82500 KiB

sequence

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

const int MXN = 100001, MXK = 202;
int N, K, s[2], e[2], stk[MXN][2], back[MXN][MXK];
ll sum[MXN], dp[MXN][2];

double cross(int x, int y, int z) {
    x = stk[x][z % 2], y = stk[y][z % 2];
    if (sum[x] == sum[y]) return -1e18;
    ll vx = -sum[N] * sum[x] + dp[x][z % 2], vy = -sum[N] * sum[y] + dp[y][z % 2];
    return 1.0 * (vy - vx) / (sum[x] - sum[y]);
}

void insert(int x, int z) {
    int zz = z % 2;
    stk[e[zz]][zz] = x;
    while (e[zz] > 1 && cross(e[zz] - 2, e[zz] - 1, z) > cross(e[zz] - 1, e[zz], z))
        stk[e[zz] - 1][zz] = stk[e[zz]][zz], e[zz]--;
    e[zz]++;
}

void query(int x, int z) {
    z--;
    int zz = z % 2;
    while (s[zz] + 1 < e[zz] && stk[s[zz]][zz] < x - 1 && cross(s[zz], s[zz] + 1, z) <= sum[x])
        s[zz]++;
    int y = stk[s[zz]][zz];
    dp[x][(z + 1) % 2] = (sum[N] - sum[x]) * (sum[x] - sum[y]) + dp[y][z % 2];
    back[x][z + 1] = y;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    cin >> N >> K; K++;
    for (int i = 1; i <= N; ++i) {
        ll n; cin >> n;
        sum[i] = sum[i - 1] + n;
    }

    insert(0, 0);
    for (int k = 1; k <= K; ++k) {
        s[(k - 1) % 2] = e[k % 2] = 0;
        for (int i = k; i <= N; ++i) {
            //dp[i][k] = max(dp[i][k], dp[j][k - 1] + (sum[N] - sum[i]) * (sum[i] - sum[j]));
            query(i, k);
            insert(i, k);
        }
    }

    cout << dp[N][K % 2] << '\n';

    int cur = back[N][K], cnt = K - 1;
    stack<int> st;
    while (cnt) {
        st.push(cur);
        cur = back[cur][cnt];
        cnt--;
    }

    while (!st.empty()) {
        cout << st.top() << ' ';
        st.pop();
    }

    cout << '\n';

    return 0;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 0 / 11
• Subtask 4: 17 / 17
• Subtask 5: 21 / 21
• Subtask 6: 29 / 29