Submission #86105

#	Time	Username	Problem	Language	Result	Execution time	Memory
86105		Shtef	Maja (COCI18_maja)	C++14	77 / 110	2063 ms	1132 KiB

maja

#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long ll;

int n, m, x1, y1, k;
ll dp[5][105][105], c[105][105], sol;
const ll inf = 1e18;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};

int main ()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin >> n >> m >> x1 >> y1 >> k;
for(int i = 1 ; i <= n ; ++i){
	for(int j = 1 ; j <= m ; ++j){
		cin >> c[i][j];
	}
}
for(int i = 1 ; i <= n ; ++i){
	for(int j = 1 ; j <= m ; ++j){
		for(int v = 0 ; v < 5 ; ++v){
			dp[v][i][j] = -inf;
		}
	}
}
for(int i = 0 ; i < 5 ; ++i){
	dp[i][x1][y1] = 0;
}
for(int i = 1 ; i <= min(k, 3 * n * m) ; ++i){
	dp[i % 5][x1][y1] = 0;
	for(int j = 1 ; j <= n ; ++j){
		for(int v = 1 ; v <= m ; ++v){
			for(int w = 0 ; w < 4 ; ++w){
				int nx = j + dx[w];
				int ny = v + dy[w];
				if(nx <= 0 || nx > n || ny <= 0 || ny > m)
					continue;
				dp[i % 5][nx][ny] = max(dp[i % 5][nx][ny], dp[(i - 1 + 5) % 5][j][v] + c[nx][ny]);
			}
		}
	}
	if(i < 4)
		continue;
	for(int j = 1 ; j <= n ; ++j){
		for(int v = 1 ; v <= m ; ++v){
			//cout << dp[i % 5][j][v] << ' ' << dp[(i - 2 + 5) % 5][j][v] << ' ' << dp[(i - 4 + 5) % 5][j][v] << ' ' << j << ' ' << v << endl;
			if(dp[i % 5][j][v] - dp[(i - 2 + 5) % 5][j][v] == dp[(i - 2 + 5) % 5][j][v] - dp[(i - 4 + 5) % 5][j][v]){
				//cout << i << " [ " << j << " ]" << "[ " << v << " ]" << endl;
				//if(2 * dp[(i - 4 + 5) % 5][j][v] + ((k - 2 * (i - 4)) / 2) * (dp[i % 5][j][v] - dp[(i - 2 + 5) % 5][j][v]) - c[j][v] > sol)
				//	cout << dp[i % 5][j][v] << ' ' << dp[(i - 2 + 5) % 5][j][v] << ' ' << i << ' ' << j << ' ' << v << endl, system("pause");
				sol = max(sol, 2 * dp[(i - 4 + 5) % 5][j][v] + ((k - 2 * (i - 4)) / 2) * (dp[i % 5][j][v] - dp[(i - 2 + 5) % 5][j][v]) - c[j][v]);
			}
		}
	}
	//system("pause");
}
cout << (k <= 3 * n * m ? dp[k % 5][x1][y1] : sol) << endl;

return 0;
}

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