Submission #1067621

#	Time	Username	Problem	Language	Result	Execution time	Memory
1067621		HorizonWest	Catfish Farm (IOI22_fish)	C++17	35 / 100	1069 ms	2097152 KiB

fish

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

#pragma GCC optimize("O3")
#define endl '\n'
#define db double
#define ll long long
#define pb push_back
#define fs first
#define sd second
#define Mod long(1e9 + 7)
#define all(x) x.begin(), x.end()
#define unvisited long(-1)
#define Eps double(1e-9)
#define _for(i, n) for(int i = 0; i < (n); i++)
#define dbg(x) cout << #x ": " << x << endl;

const int Max = 1e6 + 7, Inf = 1e9 + 7;

void print(bool x) { cout << (x ? "YES" : "NO") << endl; }

string tostring (__int128 x)
{
    string ans = "";
    while(x > 0)
    {
        ans += (x % 10 + '0');
        x /= 10;
    }
    reverse(all(ans));
    return ans;
}

long long max_weights(int n, int m, vector<int> X, vector<int> Y, vector<int> W) 
{
    //vector <vector<pair<int, int>>> v(n + 1, vector <pair<int, int>> ());

    vector <vector<ll>> v(n, vector <ll> (n+1, 0)); 
    map <int, map <int, int>> mp; 

    for(int i = 0; i < m; i++){
        v[X[i]][Y[i]+1] = W[i];
        //v[X[i]].push_back({ Y[i], W[i] });
    }

    vector <vector<ll>> dp1(n + 1, vector <ll> (n + 1, 0)), 
        dp2(n + 1, vector <ll> (n + 1, 0));

    ll ans = 0; 

    for(int i = 0; i < n; i++)
    {
        for(int j = 1; j <= n; j++){
            v[i][j] += v[i][j-1];
        }
    }

    auto C = [&] (int i, int j, int k) -> ll
    {
        ll ans = 0;
        if(i >= 0 && i < n) {
            ans = ans + max(0LL, v[i][j] - v[i][k]);
        }
        return ans;
    };

    for(int i = 0; i < n; i++)
    {   
        for(int j = 0; j <= n; j++)
        {
            for(int k = 0; k <= n; k++){
                dp1[i][j] = max(dp1[i][j], (i > 1 ? dp2[i-2][k] : 0LL) + C(i-1, j, k) + C(i+1, j, 0));
            }

            for(int k = 0; k <= n; k++){    
                dp1[i][j] = max(dp1[i][j], (i != 0 ? dp1[i-1][k] : 0) + C(i-1, j, k) + C(i+1, j, 0) - C(i, k, 0));
            }

            for(int k = 0; k <= n; k++){
                dp2[i][j] = max(dp2[i][j], (i != 0 ? dp1[i-1][k] : 0) - C(i, j, 0) + C(i+1, j, 0));
                dp2[i][j] = max(dp2[i][j], (i != 0 ? dp2[i-1][k] : 0) - C(i, j, 0) + C(i+1, j, 0));
            }

            ans = max(ans, max(dp1[i][j], dp2[i][j])); 
        }
    }

    return ans;
}

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