Submission #1101712

#	Time	Username	Problem	Language	Result	Execution time	Memory
1101712		vanea	Split the sequence (APIO14_sequence)	C++14	71 / 100	72 ms	131072 KiB

sequence

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

const int mxN = 1e5+10;
const int mxK = 205;

ll m[mxN], dp[mxN][mxK];
int last[mxN][mxK];

ld intercept(int x, int y, int k) {
    return (ld)(dp[y][k]-dp[x][k])/(ld)(m[x]-m[y]);
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<ll> v(n+1), pref(n+1);
    for(int i = 1; i <= n; i++) {
        cin >> v[i];
        pref[i] = pref[i-1]+v[i];
    }
    vector<deque<int>> q(k+1);
    q[0].push_back(0);
    for(int i = 1; i < n; i++) {
        for(int j = min(k, i); j >= 1; j--) {
            ll x = pref[n]-pref[i];
            while(q[j-1].size() >= 2 && intercept(q[j-1][0], q[j-1][1], j-1) >= x)
                q[j-1].pop_front();
            int curr = q[j-1].front();
            dp[i][j] = m[curr]*x + dp[curr][j-1] + pref[i]*x;
            m[i] = -pref[i];
            last[i][j] = curr;
            bool ok = true;
            while(q[j].size() && m[q[j].back()] == m[i]) {
                if(dp[q[j].back()][j] <= dp[i][j]) q[j].pop_back();
                else {
                    ok = false;
                    break;
                }
            }
            if(!ok) continue;
            while(q[j].size() >= 2 && intercept(q[j].back(), q[j][q[j].size()-2], j) <=
                  intercept(i, q[j][q[j].size()-2], j))
                q[j].pop_back();
            q[j].push_back(i);
        }
    }
    int idx = -1; ll mx = -1;
    for(int i = k; i < n; i++) {
        if(dp[i][k] >= mx) {
            mx = dp[i][k];
            idx = i;
        }
    }
    cout << mx << '\n';
    int prev = idx;
    vector<int> ans;
    while(k) {
        ans.push_back(prev);
        prev = last[prev][k--];
    }
    reverse(ans.begin(), ans.end());
    for(auto it : ans) {
        cout << it << ' ';
    }
    return 0;
}

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