Submission #482303

#	Time	Username	Problem	Language	Result	Execution time	Memory
482303		Blistering_Barnacles	Cloud Computing (CEOI18_clo)	C++11	36 / 100	48 ms	31076 KiB

clo

//apig's property
//Happiness can be found, even in the darkest of times, if one only remembers to turn on the light
//El Pueblo Unido Jamas Sera Vencido
//Do or do not... there is no try
//Billions of bilious blue blistering barnacles in a thundering typhoon!
#include<bits/stdc++.h>
#define fast ios_base::sync_with_stdio(0) , cin.tie(0) , cout.tie(0)
#define F first
#define S second
#define pb push_back
#define vll vector< ll >
#define vi vector< int >
#define pll pair< ll , ll >
#define pi pair< int , int >
#define all(s) s.begin() , s.end()
#define sz(s) s.size()
#define str string
#define md ((s + e) / 2)
#define mid ((l + r) / 2)
#define msdp(dp) memset(dp , -1 , sizeof dp)
#define mscl(dp) memset(dp , 0 , sizeof dp)
#define C continue
#define R return
#define B break
#define lx node * 2
#define rx node * 2 + 1
#define br(o) o ; break
#define co(o) o ; continue
#define f F.F
#define v F.S
#define t S.F
#define c S.S
using namespace std;
typedef long long ll;
ll q , dp[50005][5] , b[555555], k, l, m, n, o, p;
map < ll , ll > mp;
vll adj[555555];
const ll mod = 1e9+7;
str s;
pair < pll , pll > a[555555] ;
ll cmp(const pair < pll , pll > &a , const pair < pll , pll > &b){
    if(a.f == b.f)R a.t < b.t ;
    R a.f > b.f ;
}
void solve(){
    ll nn = 0 ;
    cin >> n ;
    o = 0 ;
    for(ll i = 1 ; i <= n ; i++){
        ll cc , ff , vv ;
        cin >> cc >> ff >> vv ;
        nn += cc ;
        a[++o] = {{ff , vv} , {1LL , cc}} ;
    }
    cin >> m ;
    for(ll i = 1 ; i <= m ; i++){
        ll cc , ff , vv ;
        cin >> cc >> ff >> vv ;
        a[++o] = {{ff , vv} , {2LL , cc}} ;
    }
    ll ans = -1e15 ;
    sort(a + 1 , a + o + 1 , cmp) ;
    for(ll j = 1 ; j <= nn ; j++){
        dp[j][0] = -1e12 ;
    }
    for(ll i = 1 ; i <= o ; i++){
        for(ll j = 0 ; j <= nn ; j++){
            dp[j][i & 1] = dp[j][!(i & 1)] ;
            if(a[i].t == 2){
                if(j + a[i].c <= nn){
                    dp[j][i & 1] = max(dp[j][i & 1] , dp[j + a[i].c][!(i & 1)] + a[i].v) ;
                }
            }
            else {
                if(j - a[i].c >= 0){
                    dp[j][i & 1] = max(dp[j][i & 1] , dp[j - a[i].c][!(i & 1)] - a[i].v) ;
                }
            }
            //cout << i << " " << j << " " << dp[j][i & 1] << " " << a[i].t << endl ;
            ans = max(ans , dp[j][i & 1]) ;
        }
    }
    cout << ans << endl ;
}
int main(){
    fast ;
    q = 1 ;
    //cin >> q ;
    while(q--){
        solve() ;
    }
}

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