Submission #1129480

#	Time	Username	Problem	Language	Result	Execution time	Memory
1129480		garam1732	Ski 2 (JOI24_ski2)	C++20	100 / 100	235 ms	281760 KiB

Main

#include <bits/stdc++.h>
using namespace std;

#define ff first
#define ss second
#define bl ' '
#define endl '\n'
#define all(v) (v).begin(), (v).end()
#define comp(v) (v).erase(unique(all(v)), (v).end())
#define lbd(v,x) lower_bound(all(v), (x))-(v).begin()
#define ubd(v,x) upper_bound(all(v), (x))-(v).begin()

typedef long long ll;
typedef pair<int, int> pi;
typedef pair<pi, int> pii;
typedef pair<int, pi> ipi;
typedef pair<pi, pi> pipi;
typedef pair<ll, ll> pll;

const int MAXN = 330;//100100*20;
const ll MOD = 1e9+7;
const ll INF = 2e18;

struct Point {
    ll h, c;
}arr[MAXN];

vector<int> v;
ll dp[MAXN][MAXN][MAXN], mn[MAXN], k;
int cnt[MAXN];

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0);

    int n; cin>>n>>k;
    for(int i=0; i<n; i++) cin>>arr[i].h>>arr[i].c;

    sort(arr, arr+n, [&](Point &a, Point &b) {
        return a.h!=b.h?a.h<b.h:a.c<b.c;
    });

    ll h = arr[0].h; mn[0] = arr[0].c; int x = 0;
    for(int i = 0, idx = 0; i < n; i++) {
        v.push_back(h);
        while(idx < n && arr[idx].h == h) x++, idx++;
        x--;
        h = max(arr[i+1].h, h+1);
    }

    mn[0] = INT_MAX;
    for(int i = 0, idx = 0; i < n; i++) {
        while(idx < n && arr[idx].h == v[i]) {
            cnt[i]++; mn[i]=min(mn[i], arr[idx].c);
            idx++;
        }

        mn[i+1] = mn[i];
    }

    memset(dp, -1, sizeof dp);
    for(int j=0; j<=n; j++) for(int k=0; k<=n; k++) dp[v.size()][j][k]=0;
    
    for(int idx=(int)v.size()-1; idx>0; idx--) {
        for(int h=0; h<=n; h++) {
            ll m = LLONG_MAX;
            for(int i=0; i<=h; i++) {
                m = min(m, dp[idx+1][cnt[idx+1]+i][h-i]+(k-mn[idx-1])*i);
                dp[idx][h][h-i] = m+mn[idx-1]*i;
            }

            for(int rem=h+1; rem<=n; rem++) dp[idx][h][rem] = dp[idx+1][cnt[idx+1]][rem];
        }
    }
    
    cout<<dp[1][cnt[1]+cnt[0]-1][1]+k*(cnt[0]-1);
}

• Subtask 1: 5 / 5
• Subtask 2: 12 / 12
• Subtask 3: 9 / 9
• Subtask 4: 33 / 33
• Subtask 5: 27 / 27
• Subtask 6: 14 / 14