Submission #206734

#TimeUsernameProblemLanguageResultExecution timeMemory
206734mode149256Dango Maker (JOI18_dango_maker)C++14
13 / 100
5 ms376 KiB
/*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 */
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