Submission #118718

#	Time	Username	Problem	Language	Result	Execution time	Memory
118718		sofhiasouza	Maja (COCI18_maja)	C++14	99 / 110	948 ms	888 KiB

maja

#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;

const int maxn = 110, maxs = 1e4+10;

int n, m, a, b, mat[maxn][maxn], k;

struct pt
{
	ll f;
	int sf;

	bool operator > (const pt &k) const{
		if(f != k.f) return f > k.f;
		return sf > k.sf;
	}

}dp[maxn][maxn][2];

pt solve()
{
	int mod = 0;
	for(int t = 1 ; t <= min(k, 10000) ; t++)
	{	
		mod ^= 1;
		for(int i = 1 ; i <= n ; i++)
		{
			for(int j = 1 ; j <= m ; j++)
			{
				if(dp[i-1][j][mod^1] > dp[i+1][j][mod^1] and dp[i-1][j][mod^1] > dp[i][j-1][mod^1] and dp[i-1][j][mod^1] > dp[i][j+1][mod^1])
				{
					dp[i][j][mod] = dp[i-1][j][mod^1];
					if(mat[i][j]+mat[i-1][j] > dp[i-1][j][mod^1].sf) dp[i][j][mod].sf = mat[i][j]+mat[i-1][j];
				}
				else if(dp[i+1][j][mod^1] > dp[i-1][j][mod^1] and dp[i+1][j][mod^1] > dp[i][j-1][mod^1] and dp[i+1][j][mod^1] > dp[i][j+1][mod^1])
				{
					dp[i][j][mod] = dp[i+1][j][mod^1];
					if(mat[i][j]+mat[i+1][j] > dp[i+1][j][mod^1].sf) dp[i][j][mod].sf = mat[i][j]+mat[i+1][j];
				}
				else if(dp[i][j-1][mod^1] > dp[i-1][j][mod^1] and dp[i][j-1][mod^1] > dp[i+1][j][mod^1] and dp[i][j-1][mod^1] > dp[i][j+1][mod^1])
				{
					dp[i][j][mod] = dp[i][j-1][mod^1];
					if(mat[i][j]+mat[i][j-1] > dp[i][j-1][mod^1].sf) dp[i][j][mod].sf = mat[i][j]+mat[i][j-1];
				}
				else
				{
					dp[i][j][mod] = dp[i][j+1][mod^1];
					if(mat[i][j]+mat[i][j+1] > dp[i][j+1][mod^1].sf) dp[i][j][mod].sf = mat[i][j]+mat[i][j+1];
				}

				if(dp[i][j][mod].f != -1) dp[i][j][mod].f += mat[i][j];
			}
		}
	}
	return dp[a][b][mod];
}

int main()
{
	cin >> n >> m >> a >> b >> k;

	for(int i = 1 ; i <= n ; i++)
	{
		for(int j = 1 ; j <= m ; j++)
		{
			cin >> mat[i][j];
		}
	}

	memset(dp, -1, sizeof dp);
	dp[a][b][0].f = 0;

	pt resp = solve();
	
	if(k < 10000) cout << resp.f << endl;
	else cout << (unsigned long long int)resp.f + (unsigned long long int)(resp.sf)*((unsigned long long int)k-10000)/2 << endl;
}

• Subtask 1: 11 / 11
• Subtask 2: 11 / 11
• Subtask 3: 11 / 11
• Subtask 4: 11 / 11
• Subtask 5: 11 / 11
• Subtask 6: 11 / 11
• Subtask 7: 11 / 11
• Subtask 8: 11 / 11
• Subtask 9: 11 / 11
• Subtask 10: 0 / 11