Submission #760988

#	Time	Username	Problem	Language	Result	Execution time	Memory
760988		drdilyor	Catfish Farm (IOI22_fish)	C++17	40 / 100	1074 ms	92496 KiB

fish

#include <bits/stdc++.h>
#include "fish.h"
using namespace std;
using ll = long long;

ll max_weights(
    int n, int m, vector<int> x, vector<int> y, vector<int> w) {
    vector arr(n+1, vector<pair<int,ll>>());
    for (int i = 0; i < m; i++) {
        arr[x[i]+1].emplace_back(y[i], w[i]);
    }
    for (int i = 0; i < n; i++)
        sort(arr[i].begin(), arr[i].end());
    auto pref = arr;
    for (int i = 0; i <= n; i++) {
        for (int j = 1; j < (int)pref[i].size(); j++) {
            pref[i][j].second += pref[i][j-1].second;
        }
    }

    auto sum_right = [&](int c, int right)->ll{
        int l = -1, r = pref[c].size();
        while (l < r-1) {
            int mid = (l+r) / 2;
            if (pref[c][mid].first <= right) l = mid;
            else r= mid;
        }
        return l==-1 ? 0 : pref[c][l].second;
    };

    auto sum = [&](int c, int l, int r) { // [l, r]
        if (l <= r) return sum_right(c, r) - sum_right(c, l-1);
        else return 0ll;
    };
    vector<vector<int>> byx(n+2), used(n+1);
    for (int i = 0; i < m; i++)
        byx[x[i]+1].push_back(y[i]);
    for (int i = 0; i <= n; i++) {
        if (i) for (int j : byx[i-1])
            used[i].push_back(j+1);
        if (n-i) for (int j : byx[i+1])
            used[i].push_back(j+1);
        used[i].push_back(0);
        sort(used[i].begin(), used[i].end());
    }


    vector memo(2, vector(n+1, vector<ll>()));
    for (int i = 1; i <= n; i++) {
        memo[0][i].assign(used[i-1].size(), -1);
        memo[1][i].assign(used[i-1].size(), -1);
    }
    
    vector memp(2, vector(n+1, vector<ll>{}));

    auto dp = [&](auto& dp, bool inc, int c, int h2_key)->ll{
        if (c == n+1) return 0;
        int h2 = used[c-1][h2_key];
        ll& res = memo[inc][c][h2_key];
        if (res!=-1) return res;
        res = 0;
        if (inc) {
            if (memp[inc][c].empty()) {
                ll mx = 0;
                for (int j = (int)used[c].size()-1; j >= 0; j--) {
                    int h3 = used[c][j];
                    ll val = sum(c-1, 0, h3-1);
                    mx = max(mx, val + max(
                                dp(dp, true, c+1, j),
                                dp(dp, false, c+1, j)));
                    memp[inc][c].push_back(mx);
                }
            }
            int start = lower_bound(used[c].begin(), used[c].end(), h2) - used[c].begin();
            for (int j = start; j < (int)used[c].size(); j++) {
                res = max(res, memp[inc][c][j]);
            }
            res -= sum(c-1, 0, h2-1);
        } else {
            if (memp[inc][c].empty()) {
                for (int j = 0; j < (int)used[c].size(); j++) {
                    int h3 = used[c][j];
                    ll val = -sum(c, 0, h3-1);
                    memp[inc][c].push_back(val + dp(dp, false, c+1, j));
                }
            }
            int finish = upper_bound(used[c].begin(), used[c].end(), h2) - used[c].begin();
            for (int j = 0; j < finish; j++) {
                int h3 = used[c][j];
                if (h3 > h2) break;
                res = max(res, memp[inc][c][j]);
            }
            res += sum(c, 0, h2-1);
        }
        res = max(res, dp(dp, true, c+1, 0));
        res = max(res, 0ll);
        return res;
    };
    ll res = dp(dp, true, 1, 0);
    return res;

}

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