Submission #1055165

#	Time	Username	Problem	Language	Result	Execution time	Memory
1055165		vjudge1	Catfish Farm (IOI22_fish)	C++17	9 / 100	47 ms	14532 KiB

fish

#include "fish.h"
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vi = vector<int>;
using vll = vector<ll>;

struct AIB {
    int n;
    vll  V;
    AIB(int N) : n(N + 2), V(N + 2, 0) {}
    void update(int p, ll v) {
        ++p;
        while(p < n) {
            V[p] = max(V[p], v);
            p += p & -p;
        }
    }

    ll query(int p) {
        ++p;
        ll re = 0;
        if(p < 0) return 0;
        while(p) {
            re = max(re, V[p]);
            p -= p & -p;
        }
        return re;
    }
};

struct maxSuf {
    int n;
    AIB A;
    maxSuf(int N) : n(N), A(N) {}

    void update(int p, ll v) { A.update(n - p - 1, v); }

    ll query(int p) { return A.query(n - p - 1); }
};


ll max_weights(int n, int m, vi X, vi Y, vi W) {
    vll DP0(m, 0), DP1(m, 0), MaTot(n, 0), Ma0(n, 0);
    vector<vi> P(n);
    for(int i = 0; i < m; ++i) {
        P[X[i]].push_back(i);
    }

    AIB MP(n);
    maxSuf MS(n);

    for(int i = 0; i < n; ++i) {
        sort(P[i].begin(), P[i].end(), [&](auto a, auto b) { return Y[a] < Y[b]; });
        if(i > 0) {
            for(auto it : P[i]) {
                DP0[it] = (i > 1 ? MaTot[i - 2] : 0) + W[it];
                DP0[it] = max(DP0[it], W[it] + MS.query(Y[it] + 1));

                MS.update(Y[it], DP0[it]);
                MaTot[i] = max(MaTot[i], DP0[it]);
                Ma0[i] = max(Ma0[i], DP0[it]);
            }
        }
        reverse(P[i].begin(), P[i].end());
        if(i + 1 < n) {
            for(auto it : P[i]) {
                DP1[it] = max((i > 1 ? MaTot[i - 2] : 0), (i ? Ma0[i - 1] : 0)) + W[it];
                DP1[it] = max(DP1[it], W[it] + MP.query(Y[it] - 1));

                MP.update(Y[it], DP1[it]);
                MaTot[i] = max(MaTot[i], DP1[it]);
            }
        }
        if(i) {
            MaTot[i] = max(MaTot[i], MaTot[i - 1]);
            Ma0[i] = max(Ma0[i], Ma0[i - 1]);
        }
    }

    return MaTot.back();
}

• Subtask 1: 0 / 3
• Subtask 2: 0 / 6
• Subtask 3: 9 / 9
• Subtask 4: 0 / 14
• Subtask 5: 0 / 21
• Subtask 6: 0 / 17
• Subtask 7: 0 / 14
• Subtask 8: 0 / 16