Submission #241109

#	Time	Username	Problem	Language	Result	Execution time	Memory
241109		DS007	Dango Maker (JOI18_dango_maker)	C++14	13 / 100	5 ms	384 KiB

dango_maker

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

signed main() {
	int n, m;
	cin >> n >> m;
	m++, n++;
	
	string s[n];
	for (int i = 1; i < n; i++)
	    cin >> s[i], s[i] = " " + s[i];
	    
	int dp[n][m][3] = {};
	for (int i = 3; i < n; i++)
	    dp[i][1][1] = dp[i - 1][1][1] + (s[i][1] == 'W' && s[i - 1][1] == 'G' && s[i - 2][1] == 'R'), dp[i][1][0] = dp[i - 1][1][1];
	for (int j = 3; j < m; j++)
	    dp[1][j][2] = dp[1][j - 1][2] + (s[1][j] == 'W' && s[1][j - 1] == 'G' && s[1][j - 2] == 'R'), dp[1][j][0] = dp[1][j - 1][2];
	    
	for (int i = 2; i < n; i++) {
	    for (int j = 2; j < m; j++) {
	        dp[i][j][0] = max(dp[i - 1][j][1], dp[i - 2][j][0]) + max({dp[i][j - 1][0], dp[i][j - 1][1], dp[i][j - 1][2]}) - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]});
	        dp[i][j][0] = max(dp[i][j][0], max(dp[i - 1][j][2], dp[i - 1][j][0]) + dp[i][j - 1][2] - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]}));
	        dp[i][j][1] = max(dp[i - 1][j][1], dp[i - 2][j][0]) + max({dp[i][j - 1][0], dp[i][j - 1][1], dp[i][j - 1][2]}) - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]}) + (s[i][j] == 'W' && s[i - 1][j] == 'G' && s[i - 2][j] == 'R');
	        dp[i][j][1] = max(dp[i][j][1], max(dp[i - 2][j][2], dp[i - 1][j][0]) + dp[i][j - 1][2] - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]}) + (s[i][j] == 'W' && s[i - 1][j] == 'G' && s[i - 2][j] == 'R'));
	        dp[i][j][2] = max(dp[i][j - 1][2], dp[i][j - 2][0]) + max({dp[i - 1][j][0], dp[i - 1][j][1], dp[i - 1][j][2]}) - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]}) + (s[i][j] == 'W' && s[i][j - 1] == 'G' && s[i][j - 2] == 'R');
	        dp[i][j][2] = max(dp[i][j][2], max(dp[i][j - 1][0], dp[i][j - 2][1]) + dp[i - 1][j][1] - max({dp[i - 1][j - 1][0], dp[i - 1][j - 1][1], dp[i - 1][j - 1][2]}) + (s[i][j] == 'W' && s[i][j - 1] == 'G' && s[i][j - 2] == 'R'));
	    }
	}
	
	cout << max({dp[n - 1][m - 1][0], dp[n - 1][m - 1][1], dp[n - 1][m - 1][2]});
	return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 0 / 20
• Subtask 3: 0 / 67