제출 #206734

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
206734		mode149256	Dango Maker (JOI18_dango_maker)	C++14	13 / 100	5 ms	376 KiB

dango_maker

/*input
3 4
RGWR
GRGG
RGWW

5 5
RGRGW
GRRGW
WGGWR
RWRGW
RGWGW

4 4
RGWR
GRRG
WGGW
WWWR



3 4
RGWR
GRGG
RGWW
*/
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef long double ld;
typedef complex<ld> cd;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;

typedef vector<int> vi;
typedef vector<vi> vii;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<vl> vll;
typedef vector<pi> vpi;
typedef vector<vpi> vpii;
typedef vector<pl> vpl;
typedef vector<cd> vcd;
typedef vector<pd> vpd;
typedef vector<bool> vb;
typedef vector<vb> vbb;
typedef std::string str;
typedef std::vector<str> vs;

#define x first
#define y second
#define debug(...) cout<<"["<<#__VA_ARGS__<<": "<<__VA_ARGS__<<"]\n"

const int MOD = 1000000007;
const ll INF = std::numeric_limits<ll>::max();
const int MX = 100101;
const ld PI = 3.14159265358979323846264338327950288419716939937510582097494L;

template<typename T>
pair<T, T> operator+(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x + b.x, a.y + b.y); }
template<typename T>
pair<T, T> operator-(const pair<T, T> &a, const pair<T, T> &b) { return pair<T, T>(a.x - b.x, a.y - b.y); }
template<typename T>
T operator*(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.x + a.y * b.y); }
template<typename T>
T operator^(const pair<T, T> &a, const pair<T, T> &b) { return (a.x * b.y - a.y * b.x); }

template<typename T>
void print(vector<T> vec, string name = "") {
	cout << name;
	for (auto u : vec)
		cout << u << ' ';
	cout << '\n';
}

using vc = vector<char>;

int main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int N, M;
	cin >> N >> M;

	vector<vc> sk(N, vc(M));

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			cin >> sk[i][j];
		}
	}

	vii idesine(N, vi(M, 0));
	vii iapacia(N, vi(M, 0));

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			if (j + 2 < M and sk[i][j] == 'R' and sk[i][j + 1] == 'G' and sk[i][j + 2] == 'W')
				idesine[i][j]++;
			if (i + 2 < N and sk[i][j] == 'R' and sk[i + 1][j] == 'G' and sk[i + 2][j] == 'W')
				iapacia[i][j]++;
		}
	}

	auto get = [&](vii vec, int i, int j) -> int {
		if (i >= 0 and j >= 0 and i < (int)vec.size() and j < (int)vec[i].size())
			return vec[i][j];
		else
			return 0;
	};

	vii sumUp(N, vi(M, 0));
	vii sumLeft(N, vi(M, 0));

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			sumUp[i][j] = get(sumUp, i - 1, j) + get(iapacia, i - 2, j);
			sumLeft[i][j] = get(sumLeft, i, j - 1) + get(idesine, i, j - 2);
		}
	}

	// for (auto u : idesine) {
	// 	for (auto e : u)
	// 		printf("%d ", e);
	// 	printf("\n");
	// }
	// printf("\n");

	// for (auto u : iapacia) {
	// 	for (auto e : u)
	// 		printf("%d ", e);
	// 	printf("\n");
	// }
	// printf("\n");

	vii trysIdesine(N, vi(M, 0));
	vii trysIapacia(N, vi(M, 0));



	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			trysIdesine[i][j] = max(
			                        get(iapacia, i, j) + get(trysIdesine, i, j - 1),
			                        get(idesine, i, j - 2) + get(idesine, i + 1, j - 2)
			                        + get(idesine, i + 2, j - 2) + get(trysIdesine, i, j - 3)
			                    );

			trysIapacia[i][j] = max(
			                        get(idesine, i, j - 2) + get(trysIapacia, i - 1, j),
			                        get(iapacia, i - 2, j) + get(iapacia, i - 2, j - 1)
			                        + get(iapacia, i - 2, j - 2) + get(trysIapacia, i - 3, j)
			                    );
		}
	}

	// for (auto u : trysIdesine) {
	// 	for (auto e : u)
	// 		printf("%d ", e);
	// 	printf("\n");
	// }
	// printf("\n");

	// for (auto u : trysIapacia) {
	// 	for (auto e : u)
	// 		printf("%d ", e);
	// 	printf("\n");
	// }
	// printf("\n");

	vii dp(N, vi(M, 0));

	for (int i = 0; i < N; ++i)
	{
		for (int j = 0; j < M; ++j)
		{
			dp[i][j] = max({
				get(dp, i, j - 3) + get(trysIapacia, i, j),
				get(dp, i - 3, j) + get(trysIdesine, i - 2, j),
				get(dp, i - 3, j - 3) + get(trysIapacia, i, j) + get(trysIdesine, i - 2, j - 3),
				get(dp, i - 3, j - 3) + get(trysIdesine, i - 2, j) + get(trysIapacia, i - 3, j),
				get(dp, i, j - 1) + get(sumUp, i, j),
				get(dp, i, j - 2) + get(sumUp, i, j) + get(sumUp, i, j - 1),
				get(dp, i - 1, j) + get(sumLeft, i, j),
				get(dp, i - 2, j) + get(sumLeft, i, j) + get(sumLeft, i - 1, j),
				get(dp, i - 1, j - 1) + get(sumLeft, i, j) + get(sumUp, i - 1, j),
				get(dp, i - 1, j - 1) + get(sumLeft, i, j - 1) + get(sumUp, i, j)
			});
		}
	}

	// for (auto u : dp) {
	// 	for (auto e : u)
	// 		printf("%d ", e);
	// 	printf("\n");
	// }
	// printf("\n");

	printf("%d\n", dp[N - 1][M - 1]);
}

/* Look for:
* special cases (n=1?)
* overflow (ll vs int?)
* the exact constraints (multiple sets are too slow for n=10^6 :( )
* array bounds
*/

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