Submission #793442

#	Time	Username	Problem	Language	Result	Execution time	Memory
793442		MISM06	Cloud Computing (CEOI18_clo)	C++14	0 / 100	146 ms	3156 KiB

clo

//0 1 1 0 1
//0 1 0 0 1
//1 0 0 1 1
//0 1 1 0 1
#include <bits/stdc++.h>
#pragma GCC optimize("Ofast,unroll-loops")
// #pragma GCC target("avx2")

using namespace std;

#define F 			first
#define S 			second
#define pb 			push_back
#define sze			size()
#define	all(x)		x.begin() , x.end()
#define wall__		cout << "--------------------------------------\n";
#define kids		int mid = (tl + tr) >> 1, cl = v << 1, cr = v << 1 | 1
#define file_io		freopen("input.cpp", "r", stdin); freopen("output.cpp", "w", stdout);

typedef long long ll;
typedef long double dl;
typedef pair < int , int > pii;
typedef pair < int , ll > pil;
typedef pair < ll , int > pli;
typedef pair < ll , ll > pll;
typedef pair < int , pii > piii;
typedef pair < ll, pll > plll;


const ll N = 2000 + 10;
const ll mod = 1e9 + 7;
const ll inf = 2e16;
const ll INF = 1e9 + 10;
const ll lg = 32;

ll dp[2][N][100], n, m, c1[N], f1[N], v1[N], c2[N], f2[N], v2[N]; 

vector < piii > o1, o2;

void solve () {

	cin >> n;
	for (int i = 1; i <= n; i++) {
		int c, f, v;
		cin >> c >> f >> v;
		o1.pb({f, {c, v}});
	}
	cin >> m;
	for (int i = 1; i <= m; i++) {
		int c, f, v; cin >> c >> f >> v;
		o2.pb({f, {c, v}});
	}
	sort(all(o1));
	sort(all(o2));
	for (int i = 1; i <= n; i++) {
		f1[i] = o1[i - 1].F; c1[i] = o1[i - 1].S.F; v1[i] = o1[i - 1].S.S;
	}
	for (int i = 1; i <= m; i++) {
		f2[i] = o2[i - 1].F; c2[i] = o2[i - 1].S.F; v2[i] = o2[i - 1].S.S;
	}
	int t = 0;
	for (int i = 0; i <= n; i++) {
		t ^= 1;
		for (int j = 1; j <= m; j++) {
			for (int k = 0; k <= 50; k++) {
				dp[t][j][k] = 0;
				if (i == 0) {
					if (k >= c2[j]) dp[t][j][k] = max(dp[t][j - 1][k - c2[j]] + v2[j], dp[t][j][k]);
					dp[t][j][k] = max(dp[t][j][k], dp[t][j - 1][k]);
				} else {
					dp[t][j][k] = max({0ll, dp[t][j - 1][k], dp[t^1][j][k]});
					if (f1[i] >= f2[j]) dp[t][j][k] = max(dp[t][j][k], dp[t^1][j][k + c1[i]] - v1[i]);
					if (k >= c2[j]) dp[t][j][k] = max(dp[t][j][k], dp[t^1][j - 1][k - c2[j]] + v2[j]);
				}
			}
		}
	}
	cout << dp[t][m][0] << '\n';
	
}


int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int t = 1;
	// cin >> t;
	while (t--) {solve();}
    return 0;
}
/*
*/
//shrek is love;

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