Submission #417080

#	Time	Username	Problem	Language	Result	Execution time	Memory
417080		egekabas	Cloud Computing (CEOI18_clo)	C++14	54 / 100	1872 ms	3864 KiB

clo

#include <bits/stdc++.h>
#define all(x) (x).begin(), (x).end()
#define ff first
#define ss second
#define pb push_back
#define mp make_pair
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<ull, ull> pull;
typedef pair<ll, ll> pii;
typedef pair<ld, ld> pld;
const ll inf = 1e17;
struct st{
    ll c, f, v;
};
int sfst = 0;
ll sf(st a, st b){
    if(a.f != b.f)
        return a.f > b.f;
    if(sfst)
        return a.v < b.v;
    else
        return a.v > b.v;
}
ll n, m;
st a[2009];
st b[2009];
ll dp[2001][109];
ll nxt[2001][109];
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    
    cin >> n;
    for(ll i = 0; i < n; ++i)
        cin >> a[i].c >> a[i].f >> a[i].v;
    cin >> m;
    for(ll i = 0; i < m; ++i)
        cin >> b[i].c >> b[i].f >> b[i].v;
    
    sort(a, a+n, sf); sfst = 1;
    sort(b, b+m, sf);
    
    for(ll j = 0; j <= m; ++j)
        for(ll k = 0; k <= 55; ++k)
            dp[j][k] = nxt[j][k] = -inf;
    dp[0][0] = 0;
    ll ans = 0;
    
    for(ll i = 0; i < n; ++i){
        for(ll j = 0; j < m; ++j)
            for(ll k = 0; k <= 55; ++k){
                nxt[j][k] = max(nxt[j][k], dp[j][k]);
                dp[j+1][k] = max(dp[j+1][k], dp[j][k]);
                if(k >= b[j].c)
                    dp[j+1][k-b[j].c] = max(dp[j+1][k-b[j].c], dp[j][k]+b[j].v);
                if(a[i].f >= b[j].f){
                    if(k+a[i].c <= 55){
                        nxt[j][k+a[i].c] = max(nxt[j][k+a[i].c], dp[j][k]-a[i].v);
                    }
                    if(k+a[i].c-b[j].c >= 0 && k+a[i].c-b[j].c <= 55)
                        nxt[j+1][k+a[i].c-b[j].c] = max(nxt[j+1][k+a[i].c-b[j].c], dp[j][k]-a[i].v+b[j].v);
                }
            }
        for(ll j = 0; j <= m; ++j)
            for(ll k = 0; k <= 55; ++k){
                ans = max(ans, nxt[j][k]);
                ans = max(ans, dp[j][k]);
                
                dp[j][k] = nxt[j][k];
                nxt[j][k] = -inf;
            }
    }
    
    cout << ans << '\n';
}

• Subtask 1: 0 / 18
• Subtask 2: 18 / 18
• Subtask 3: 18 / 18
• Subtask 4: 0 / 18
• Subtask 5: 18 / 18
• Subtask 6: 0 / 10