Submission #203789

#TimeUsernameProblemLanguageResultExecution timeMemory
203789dimash241Two Dishes (JOI19_dishes)C++17
10 / 100
1181 ms168832 KiB
//#pragma GCC target("avx2") //#pragma GCC optimize("O3") //# include <x86intrin.h> # include <bits/stdc++.h> # include <ext/pb_ds/assoc_container.hpp> # include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; using namespace std; template<typename T> using ordered_set = tree <T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>; #define _USE_MATH_DEFINES_ #define ll long long #define ld long double #define Accepted 0 #define pb push_back #define mp make_pair #define sz(x) (int)(x.size()) #define every(x) x.begin(),x.end() #define F first #define S second #define lb lower_bound #define ub upper_bound #define For(i,x,y) for (ll i = x; i <= y; i ++) #define FOr(i,x,y) for (ll i = x; i >= y; i --) #define SpeedForce ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0) // ROAD to... Red inline void Input_Output () { //freopen(".in", "r", stdin); //freopen(".out", "w", stdout); } const double eps = 0.000001; const ld pi = acos(-1); const int maxn = 1e7 + 9; const int mod = 1e9 + 7; const ll MOD = 1e18 + 9; const ll INF = 1e18 + 123; const int inf = 2e9 + 11; const int mxn = 1e6 + 9; const int N = 1e6 + 123; const int M = 22; const int pri = 997; const int Magic = 2101; const int dx[] = {-1, 0, 1, 0}; const int dy[] = {0, -1, 0, 1}; int n, m; struct css { ll s, t, p; } a[N], b[N]; ll dp[2020][2020]; int main () { SpeedForce; cin >> n >> m; for (int i = 1; i <= n; i ++) { cin >> a[i].s >> a[i].t >> a[i].p; a[i].s += a[i-1].s; } for (int i = 1; i <= m; i ++) { cin >> b[i].s >> b[i].t >> b[i].p; b[i].s += b[i-1].s; } for (int i = 0; i <= n; i ++) { for (int j = 0; j <= m; j++) dp[i][j] = -INF * (i + j > 0); } for (int i = 0; i <= n; i ++) { // cout << "for : " << i << '\n'; for (int j = 0; j <= m; ++j) { if (i < n) { dp[i+1][j] = max(dp[i+1][j], dp[i][j] + (a[i+1].s + b[j].s <= a[i+1].t ? a[i+1].p : 0)); } if (j < m) { dp[i][j+1] = max(dp[i][j+1], dp[i][j] + (a[i].s + b[j+1].s <= b[j+1].t ? b[j+1].p : 0)); } // cout << j << ' ' << dp[i][j] << '\n'; } } cout << dp[n][m] << '\n'; return Accepted; } // B...a
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