제출 #417534

#제출 시각아이디문제언어결과실행 시간메모리
417534AmineWeslatiDango Maker (JOI18_dango_maker)C++14
0 / 100
69 ms142148 KiB
//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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