이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
//Never stop trying
#include "bits/stdc++.h"
using namespace std;
#define boost ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0)
typedef long long ll;
typedef string str;
typedef long double ld;
typedef pair<int, int> pi;
#define fi first
#define se second
typedef vector<int> vi;
typedef vector<pi> vpi;
#define pb push_back
#define eb emplace_back
#define sz(x) (int)x.size()
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define endl "\n"
#define FOR(i,a,b) for (int i = (a); i < (b); ++i)
#define ROF(i,a,b) for (int i = (b)-1; i >= (a); --i)
const int MOD = 1e9 + 7; //998244353
const ll INF = 1e18;
const int nx[4] = {0, 0, 1, -1}, ny[4] = {1, -1, 0, 0}; //right left down up
template<class T> using V = vector<T>;
template<class T> bool ckmin(T& a, const T& b) { return a > b ? a = b, 1 : 0; }
template<class T> bool ckmax(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
ll cdiv(ll a, ll b) { return a / b + ((a ^ b) > 0 && a % b); } // divide a by b rounded up
//constexpr int log2(int x) { return 31 - __builtin_clz(x); } // floor(log2(x))
mt19937 rng(chrono::system_clock::now().time_since_epoch().count());
//mt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());
ll random(ll a, ll b){
return a + rng() % (b - a + 1);
}
#ifndef LOCAL
#define cerr if(false) cerr
#endif
#define dbg(x) cerr << #x << " : " << x << endl;
#define dbgs(x,y) cerr << #x << " : " << x << " / " << #y << " : " << y << endl;
#define dbgv(v) cerr << #v << " : " << "[ "; for(auto it : v) cerr << it << ' '; cerr << ']' << endl;
#define here() cerr << "here" << endl;
void IO() {
#ifdef LOCAL
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
}
/////////////////////////ONLY CLEAN CODES ALLOWED/////////////////////////
const int MX=3010;
int N,M,g[MX][MX];
int memo[MX][MX][2][2];
int solve(int x, int y, int m=0, int m2=0){
if(x<0 || y>=M) return 0;
int &ind=memo[x][y][m][m2];
if(ind!=-1) return ind;
int ans=solve(x-1,y+1,m2,0);
//down
if(x+2<N && g[x][y]==0 && g[x+1][y]==1 && g[x+2][y]==2){
if(m+m2==0) ckmax(ans,solve(x-1,y+1,m2,0)+1);
}
//right
if(y+2<M && g[x][y]==0 && g[x][y+1]==1 && g[x][y+2]==2){
ckmax(ans,solve(x-1,y+1,m2,1)+1);
}
return ind=ans;
}
int main() {
boost; IO();
cin>>N>>M;
FOR(i,0,N) FOR(j,0,M){
char c; cin>>c;
if(c=='R') g[i][i]=0;
else if(c=='G') g[i][j]=1;
else g[i][j]=2;
}
memset(memo,-1,sizeof(memo));
int ans=0;
FOR(i,0,N) ans+=solve(i,0);
FOR(i,1,M) ans+=solve(N-1,i);
cout << ans << endl;
return 0;
}
//Change your approach
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