Submission #796667

#	Time	Username	Problem	Language	Result	Execution time	Memory
796667		adrilen	메기 농장 (IOI22_fish)	C++17	14 / 100	1062 ms	13724 KiB

fish

#include "fish.h"
#include<bits/stdc++.h>
using namespace std;
using ll = long long;
typedef pair<int, int> pii;
using arr = array<ll, 2>;
constexpr int maxn = 305, maxm = 3e5;
int n, m;


vector <arr> fish[maxn];

ll dp[maxn][3][maxn] = { 0 };

long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y, std::vector<int> W) {
    n = N, m = M;

    for (int &i: X) i++;
    for (int &i : Y) i++;
    for (int i = 0; i < maxn; i++) fish[i].push_back({0, 0});
    for (int i = 0; i < m; i++)
    {
        fish[X[i]].push_back(arr{Y[i], W[i]});
    }

    
    for (int i = 0; i <= n; i++) sort(fish[i].begin(), fish[i].end());

    for (int i = 1; i <= n; i++) {
        for (int y = 1; y < (int)fish[i].size(); y++)
        {
            fish[i][y][1] += fish[i][y - 1][1];
        }
    }


    // Case 0: du er på bunnen
    // Case 1: stigende

    ll output = 0;

    for (int x = 2; x <= n + 1; x++)
    {
        for (int y = 0; y <= n; y++)
        {
            /*
            Case 0
            */

            // Case 0, 0 (y er her forrige i stedet for kommende)
            for (int y = 0; y <= n; y++)
            {
                dp[x][0][0] = max(dp[x][0][0], dp[x - 1][0][y]);
            }

            dp[x][0][y] = max(
                dp[x - 1][1][y],
                dp[x - 1][2][y]
            ) + (*(upper_bound(fish[x].begin(), fish[x].end(), arr{y + 1, -1}) - 1))[1]; // Mengden fisk


            /*
            Case 1
            Stigende
            */

            for (int a = 1; a <= y; a++)
            {
                dp[x][1][y] = max(
                    dp[x][1][y], 
                    dp[x - 1][1][a] + 
                    (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{y + 1, -1}) - 1))[1] -
                    (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{a + 1, -1}) - 1))[1]
                );
            }

            for (int a = 0; a <= n; a++) {
                dp[x][1][y] = max(
                    dp[x][1][y], 
                    dp[x - 1][0][a] + 
                    max((*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{y + 1, -1}) - 1))[1] -
                    (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{a + 1, -1}) - 1))[1], 0ll)
                );
            }


            /*
            Case 2
            Fallende
            */

            for (int a = y + 1; a <= n; a++)
            {
                dp[x][2][y] = max(
                    dp[x][2][y],
                    max(dp[x - 1][1][a], dp[x - 1][2][a]) +
                    (*(upper_bound(fish[x].begin(), fish[x].end(), arr{a + 1, -1}) - 1))[1] -
                    (*(upper_bound(fish[x].begin(), fish[x].end(), arr{y + 1, -1}) - 1))[1]
                );
            }

        }



    }

    // for (int x = 0; x <= n + 1; x++)
    // {   
    //     cout << "x: " << x << "\n";
    //     for (int y = 0; y < 3; y++)
    //     {
    //         for (int c = 0; c <= n; c++) cout << dp[x][y][c] << " ";
    //         cout << "\n";
    //     }
    //     cout << "\n";
    // }


    for (int i = 0; i <= n; i++) output = max(output, dp[n + 1][0][i]);

    return output;
}

/*
        for (int a = 0; a <= n; a++) // Gammel
        {
            
            // Case 1: stigende
            // Forrige må være case 0, eller også stigende

            for (int y = 0; y <=n; y++)
            {
                dp[x][1][y] = max(dp[x][1][y], 
                dp[x - 1][0][a] + 
                (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{y + 1, -1})))[1]
                - (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{a + 1, -1})))[1]
                );
            }

            for (int y = a; y <= n; y++)
            {
                dp[x][1][y] = max(dp[x][1][y], 
                dp[x - 1][1][a] + 
                (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{y + 1, -1})))[1]
                - (*(upper_bound(fish[x - 1].begin(), fish[x - 1].end(), arr{a + 1, -1})))[1]
                );
            }

            // Case 2: fallende
            // Forrige må ha vært stigende eller fallende

            // Fallende
            for (int y = 1; y < a; y++)
            {
                dp[x][2][y] = max(dp[x][2][y], 
                dp[x - 1][2][a] - (*(upper_bound(fish[x].begin(), fish[x].end(), arr{y + 1, -1})))[1]
                + (*(upper_bound(fish[x].begin(), fish[x].end(), arr{a + 1, -1})))[1]
                );
            }

            // stigende
            for (int y = a; y <= n; y++) {
                dp[x][2][y] = max(dp[x][2][y], 
                dp[x - 1][1][a] + (*(upper_bound(fish[x].begin(), fish[x].end(), arr{y + 1, -1})))[1]
                + (*(upper_bound(fish[x].begin(), fish[x].end(), arr{a + 1, -1})))[1]
                );
            }

            if (x == 1) break;
        }

*/

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