Submission #731277

#	Time	Username	Problem	Language	Result	Execution time	Memory
731277		alextodoran	메기 농장 (IOI22_fish)	C++17	0 / 100	113 ms	12040 KiB

fish

/**
 _  _   __  _ _ _  _  _ _
 |a  ||t  ||o    d | |o  |
| __    _| | _ | __|  _ |
| __ |/_  | __  /__\ / _\|

**/

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const int N_MAX = 100000;
const int M_MAX = 300000;
const ll INF = LLONG_MAX / 3;

int N, M;
int X[M_MAX + 2], Y[M_MAX + 2], W[M_MAX + 2];

vector <int> onX[N_MAX + 2];

ll Fen[N_MAX + 2];
vector <int> updi;
void init () {
    while (updi.empty() == false) {
        Fen[updi.back()] = -INF;
        updi.pop_back();
    }
}
void update (int pos, ll val) {
    for (int i = pos; i <= N; i += i & -i) {
        updi.push_back(i);
        Fen[i] = max(Fen[i], val);
    }
}
ll query (int pos) {
    ll val = -INF;
    for (int i = pos; i >= 1; i -= i & -i) {
        val = max(val, Fen[i]);
    }
    return val;
}

ll dpUp[M_MAX + 2];
ll dpDown[M_MAX + 2];
ll fullUp[M_MAX + 2];
ll fullDown[M_MAX + 2];

ll max_weights (int _N, int _M, vector <int> _X, vector <int> _Y, vector <int> _W) {
    N = _N; M = _M;
    for (int i = 1; i <= M; i++) {
        X[i] = _X[i - 1] + 1;
        Y[i] = _Y[i - 1] + 1;
        W[i] = _W[i - 1];
        onX[X[i]].push_back(i);
    }
    fill(Fen + 1, Fen + N + 1, -INF);
    fullUp[0] = -INF;
    for (int x = 1; x <= N; x++) {
        sort(onX[x].begin(), onX[x].end(), [&] (const int &i, const int &j) {
            return Y[i] < Y[j];
        });
        init();
        update(1, fullDown[x]);
        for (int i : onX[x - 1]) {
            update(Y[i], dpUp[i]);
        }
        for (int i : onX[x]) {
            dpUp[i] = query(Y[i] - 1) + W[i];
            update(Y[i], dpUp[i]);
        }
        fullUp[x + 1] = max({fullUp[x], fullDown[x], query(N)});
        reverse(onX[x].begin(), onX[x].end());
        init();
        update(1, fullUp[x - 1]);
        for (int i : onX[x - 1]) {
            update((N - Y[i] + 1), dpDown[i]);
        }
        for (int i : onX[x]) {
            dpDown[i] = query((N - Y[i] + 1) - 1) + W[i];
            update((N - Y[i] + 1), dpDown[i]);
        }
        fullDown[x + 1] = max({fullUp[x], fullDown[x], query(N)});
        reverse(onX[x].begin(), onX[x].end());
    }
    return max(fullDown[N + 1], fullUp[N]);
}

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