Submission #563878

#	Time	Username	Problem	Language	Result	Execution time	Memory
563878		maomao90	Two Dishes (JOI19_dishes)	C++17	15 / 100	166 ms	32048 KiB

dishes

// Hallelujah, praise the one who set me free
// Hallelujah, death has lost its grip on me
// You have broken every chain, There's salvation in your name
// Jesus Christ, my living hope
#include <bits/stdc++.h> 
using namespace std;

template <class T>
inline bool mnto(T& a, T b) {return a > b ? a = b, 1 : 0;}
template <class T>
inline bool mxto(T& a, T b) {return a < b ? a = b, 1: 0;}
#define REP(i, s, e) for (int i = s; i < e; i++)
#define RREP(i, s, e) for (int i = s; i >= e; i--)
typedef long long ll;
typedef long double ld;
#define FI first
#define SE second
typedef pair<int, int> ii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> iii;
#define ALL(_a) _a.begin(), _a.end()
#define pb push_back
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<ii> vii;
typedef vector<iii> viii;

#ifndef DEBUG
#define cerr if (0) cerr
#endif

const int INF = 1000000005;
const ll LINF = 1000000000000000005ll;
const int MAXN = 1000005;

int n, m;
ll a[MAXN], s[MAXN], p[MAXN], c[MAXN];
ll b[MAXN], t[MAXN], q[MAXN], d[MAXN];
ll ans;

int main() {
#ifndef DEBUG
    ios::sync_with_stdio(0), cin.tie(0);
#endif
    cin >> n >> m;
    REP (i, 1, n + 1) {
        cin >> a[i] >> s[i] >> p[i];
        c[i] = a[i] + c[i - 1];
    }
    ll sm = 0, res = 0;
    REP (i, 1, m + 1) {
        cin >> b[i] >> t[i] >> q[i];
        d[i] = b[i] + d[i - 1];
        sm += b[i];
        res += q[i];
    }
    ans = -LINF;
    if (n <= 2000 && m <= 2000) {
        vector<vll> dp(n + 5, vll(m + 5, -LINF));
        REP (i, 0, n + 1) {
            REP (j, 0, m + 1) {
                if (i == 0 && j == 0) {
                    dp[i][j] = 0;
                }
                if (i) {
                    mxto(dp[i][j], dp[i - 1][j] + p[i] * (d[j] <= s[i] - c[i]));
                }
                if (j) {
                    mxto(dp[i][j], dp[i][j - 1] + q[j] * (c[i] <= t[j] - d[j]));
                }
            }
        }
        ans = dp[n][m];
    } else {
        int j = m;
        ll gt = t[1];
        REP (i, 0, n + 1) {
            sm += a[i];
            res += p[i];
            while (j > 0 && sm > gt) {
                sm -= b[j];
                res -= q[j--];
            }
            if (sm > gt) continue;
            mxto(ans, res);
        }
    }
    cout << ans << '\n';
    return 0;
}

• Subtask 1: 5 / 5
• Subtask 2: 3 / 3
• Subtask 3: 7 / 7
• Subtask 4: 0 / 39
• Subtask 5: 0 / 11
• Subtask 6: 0 / 9
• Subtask 7: 0 / 17
• Subtask 8: 0 / 9