Submission #431314

#	Time	Username	Problem	Language	Result	Execution time	Memory
431314		Collypso	Boxes with souvenirs (IOI15_boxes)	C++17	25 / 100	7 ms	8140 KiB

boxes

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vt vector
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define sz(x) (int) (x).size()
#pragma GCC optimize ("O3")
#pragma GCC optimize ("O2")
#define F first
#define S second
//#define endl '\n'
//#define int long long

using namespace std;

ll delivery(int N, int K, int L, int p[])
{
    vt<pii> v;
    for(int i = 0; i < N; i++) v.pb({p[i], i});
    sort(all(v));
    vt<int> order(N);
    for(int i = 0; i < N; i++) order[i] = v[i].S;

    ll mn = LLONG_MAX;
    ll dp[N][K + 1];
    dp[0][K] = 2ll * min(p[order[0]], L - p[order[0]]);
    for(int i = K - 1; i >= 0; i--)
        dp[0][i] = min(p[order[0]], L - p[order[0]]);
    for(int i = 1; i < N; i++)
    {
        ll tmp = (L - max(p[order[i]], p[order[i - 1]]));
        tmp += min(p[order[i]], p[order[i - 1]]);
        ll dist = min((ll) abs(p[order[i]] - p[order[i - 1]]), tmp);

        dp[i][K] = dp[i - 1][1] + dist + min(p[order[i]], L - p[order[i]]);
        dp[i][K - 1] = dp[i - 1][K] + min(p[order[i]], L - p[order[i]]);

        for(int j = K - 2; j >= 0; j--)
            dp[i][j] = dp[i - 1][j + 1] + dist;
    }

    ll cur_mn = LLONG_MAX;
    for(int i = 0; i <= K; i++) cur_mn = min(cur_mn, dp[N - 1][i]);

    if (cur_mn + min(p[order[N - 1]], L - p[order[N - 1]]) <= mn)
    {
        mn = min(mn, cur_mn + min(p[order[N - 1]], L - p[order[N - 1]]));
        //cout << mn << endl;
        //for(int x : order) cout << x << " ";
        //cout << endl;
        //for(int x : order) cout << p[x] << " ";
        //cout << endl;
        for(int i = K; i >= 0; i--)
        {
            //for(int j = 0; j < N; j++) cout << dp[j][i] << " ";
            //cout << endl;
        }
        //cout << endl;

    }

    //for(int x : order) cout << p[x] << " ";
    //cout << endl;
    for(int i = K; i >= 0; i--)
    {
        //for(int j = 0; j < N; j++) cout << dp[j][i] << " ";
        //cout << endl;
    }
        //cout << endl;
    return mn;
}

• Subtask 1: 0 / 10
• Subtask 2: 10 / 10
• Subtask 3: 15 / 15
• Subtask 4: 0 / 15
• Subtask 5: 0 / 20
• Subtask 6: 0 / 30