Submission #1279561

#	Time	Username	Problem	Language	Result	Execution time	Memory
1279561		hiepsimauhong	Cloud Computing (CEOI18_clo)	C++20	100 / 100	233 ms	1340 KiB

clo

#include <bits/stdc++.h>

using namespace std;

using ll = long long;
const int64_t oo = 1e15;
#define int long long

#define quickly ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define FOR(I, L, R) for(int I(L) ; I <= (int)R ; ++I)
#define FOD(I, R, L) for(int I(R) ; I >= (int)L ; --I)
#define FOA(I, A) for(auto I : A)
#define print(A,L,R) FOR(OK, L, R){if(A[OK]<=-oo / 10||A[OK]>=oo)cout<<"- ";else cout<<A[OK]<<' ';}cout<<'\n';
#define prints(A) FOA(OK, A){cout<<OK<<' ';}cout << '\n';
#define printz(A,L,R) FOR(OK, 0, L){FOR(KO, 0, R){if(A[OK][KO]>-oo&&A[OK][KO]<oo)cout<<A[OK][KO]<<' ';else cout << "- ";} cout << '\n';}cout << '\n';
#define fs first
#define sd second
#define th second.second
#define ii pair<int,int>
#define iii pair<int, ii>
#define all(A) A.begin(), A.end()

const int N = 4000 + 5;
const int mod = 1e9 + 7;

struct item{
        int wei, cap, val, in;
};

int n, m, s1, s2, lim;
item a[N];
int dp[25 * N];

signed main(){ quickly
        cin >> n;

        FOR(i, 1, n){
                cin >> a[i].wei >> a[i].cap >> a[i].val;
                a[i].in = i;
                s1 += a[i].wei;
        }

        cin >> m;

        FOR(i, n + 1, m + n){
                cin >> a[i].wei >> a[i].cap >> a[i].val;
                a[i].in = i;
                s2 += a[i].wei;
        }

        lim = max(s1, s2);

        sort(a + 1, a + 1 + n + m, [&](item &u, item &v){
                if(u.cap == v.cap){
                        return u.in < v.in;
                }
                return u.cap > v.cap;
        });

        memset(dp, -0x3f, sizeof dp);
        dp[0] = 0;

        FOR(i, 1, n + m){
                int w = a[i].wei, v = a[i].val;

                if(a[i].in <= n){
                        FOD(s, lim, w){
                                dp[s] = max(dp[s], dp[s - w] - v);
                        }
                }
                else{
                        FOR(s, 0, lim - w){
                                dp[s] = max(dp[s], dp[s + w] + v);
                        }
                }
        }

        cout << *max_element(dp, dp + 1 + lim);
}

• Subtask 1: 18 / 18
• Subtask 2: 18 / 18
• Subtask 3: 18 / 18
• Subtask 4: 18 / 18
• Subtask 5: 18 / 18
• Subtask 6: 10 / 10