Submission #1056454

#	Time	Username	Problem	Language	Result	Execution time	Memory
1056454		aykhn	Catfish Farm (IOI22_fish)	C++17	0 / 100	94 ms	16732 KiB

fish

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

using namespace std;

const int MXN = 3e2 + 5;

long long dp[2][MXN][MXN], val[MXN][MXN];
long long lft[MXN][MXN], rig[MXN][MXN], in[MXN][MXN];

/*
0 = len[i] < len[i + 1];
1 = len[i] > len[i + 1];
*/
long long max_weights(int N, int M, vector<int> X, vector<int> Y, vector<int> W) 
{
    for (int i = 0; i < M; i++) val[X[i] + 1][Y[i] + 1] = 1LL * val[X[i] + 1][Y[i] + 1] + 1LL * W[i];
    for (int i = 1; i <= N; i++) 
    {
        for (int j = 1; j <= N; j++)
        {
            lft[i][j] = lft[i][j - 1] + val[i - 1][j];
            rig[i][j] = rig[i][j - 1] + val[i + 1][j];
            in[i][j] = in[i][j - 1] + val[i][j];
        }
    }
    for (int i = 1; i <= N; i++)
    {
        for (int j = 1; j <= N; j++)
        {
            dp[0][i][j] = dp[1][i][j] = lft[i][j] + rig[i][j];
            if (i == 1) continue;
            for (int k = 1; k <= N; k++)
            {
                if (k <= j)
                {
                    long long res = (dp[0][i - 1][k] - rig[i - 1][k]) + ((lft[i][j] - in[i - 1][k]) + rig[i][j]);
                    dp[0][i][j] = max(dp[0][i][j], res);
                }
                if (k >= j)
                {
                    long long res = dp[1][i - 1][k] - in[i][j] + rig[i][j];
                    dp[1][i][j] = max(dp[1][i][j], res);
                }
            }
            if (i == 2) continue;
            for (int k = 1; k <= N; k++)
            {
                long long res = max(dp[0][i - 2][k], dp[1][i - 2][k]) + lft[i][j] + rig[i][j];
                if (k <= j) res -= rig[i - 2][k];
                else res -= lft[i][j];
                dp[0][i][j] = max(dp[0][i][j], res);
                dp[1][i][j] = max(dp[1][i][j], res);
            }
        }
    }
    long long ans = 0;
    for (int i = 1; i <= N; i++) 
    {
        for (int j = 1; j <= N; j++) 
        {
            ans = max({ans, dp[0][i][j], dp[1][i][j]});
        }
    }
    return ans;
}

• 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