Submission #260477

#	Time	Username	Problem	Language	Result	Execution time	Memory
260477		tbzard	Holiday (IOI14_holiday)	C++14	47 / 100	1049 ms	5428 KiB

holiday

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

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<int, pii> pipii;
typedef pair<pii, int> piipi;
typedef pair<pii, pii> piipii;

#define mp make_pair
#define fi first
#define se second
#define all(a) (a).begin(), (a).end()
#define sz(a) (int)(a).size()
#define eb emplace_back

ll findMaxAttraction(int n, int start, int d, int attraction[]){
    long long ans = 0;
    if(n <= 3000){
        for(int l=0;l<n;l++){
            multiset<int> mx1, mx2;
            ll res = 0;
            for(int r=l;r<n;r++){
                int cost = 0;
                if(l <= start && start <= r) cost = min(abs(start-l + (r-l)), abs(r-start + (r-l)));
                else if(r <= start) cost = start-l;
                else cost = r-start;

                if(d-cost < 0) continue;

                mx1.insert(attraction[r]);
                while(sz(mx2) < d-cost){
                    if(mx1.empty()) break;
                    int val = *mx1.rbegin();
                    mx1.erase(--mx1.end());
                    res += val;
                    mx2.insert(val);
                }

                while(sz(mx2) > d-cost){
                    int val = *mx2.begin();
                    mx2.erase(mx2.begin());
                    res -= val;
                    mx1.insert(val);
                }

                while(!mx1.empty() && !mx2.empty() && *mx1.rbegin() > *mx2.begin()){
                    int val1 = *mx1.rbegin(), val2 = *mx2.begin();
                    mx1.erase(--mx1.end());
                    mx2.erase(mx2.begin());
                    res -= val2;
                    res += val1;
                    mx1.insert(val2);
                    mx2.insert(val1);
                }
                ans = max(ans, res);
            }
        }
        // for(int i=1;i<(1<<n);i++){
        //     vector<int> b;
        //     int l = 1e9, r = 0;
        //     for(int j=0;j<n;j++){
        //         if((i>>j)&1){
        //             b.eb(j);
        //             l = min(l, j);
        //             r = max(r, j);
        //         }
        //     }
        //     int cost = 0;
        //     if(l <= start && start <= r) cost = min(abs(start-l + (r-l)), abs(r-start + (r-l)));
        //     else if(r <= start) cost = start-l;
        //     else cost = r-start;

        //     if(cost > d) continue;
        //     vector<int> c;
        //     for(int i=0;i<sz(b);i++){
        //         int val = attraction[b[i]];
        //         c.eb(val);
        //     }
        //     sort(all(c));
        //     reverse(all(c));
        //     ll res = 0;
        //     int cnt = 0;
        //     for(int i=0;i<sz(c);i++){
        //         if(cnt < d-cost){
        //             cnt++;
        //             res += c[i];
        //         }
        //     }
        //     ans = max(ans, res);
        // }
    }
    else if(start == 0){
        multiset<int> mx1, mx2;
        long long res = 0;
        for(int i=0;i<n;i++){
            mx1.insert(attraction[i]);

            while(sz(mx2) < d){
                if(mx1.empty()) break;
                int val = *mx1.rbegin();
                mx1.erase(--mx1.end());
                res += val;
                mx2.insert(val);
            }

            while(sz(mx2) > d){
                int val = *mx2.begin();
                mx2.erase(mx2.begin());
                res -= val;
                mx1.insert(val);
            }

            while(!mx1.empty() && !mx2.empty() && *mx1.rbegin() > *mx2.begin()){
                int val1 = *mx1.rbegin(), val2 = *mx2.begin();
                mx1.erase(--mx1.end());
                mx2.erase(mx2.begin());
                res -= val2;
                res += val1;
                mx1.insert(val2);
                mx2.insert(val1);
            }
            ans = max(ans, res);

            d--;
            if(d < 0) break;
        }
    }
    return ans;
}

• Subtask 1: 7 / 7
• Subtask 2: 23 / 23
• Subtask 3: 17 / 17
• Subtask 4: 0 / 53