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| #include <bits/stdc++.h>
#pragma optimization_level 3
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math,O3")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
using namespace std;
struct PairHash {
inline std::size_t operator()(const std::pair<int, int> &v) const { return v.first * 31 + v.second; }
};
#define Code ios_base::sync_with_stdio(false);
#define By ios::sync_with_stdio(0);
#define Sumfi cout.tie(NULL);
using ll = long long;
using ld = long double;
const ld PI = 3.14159265358979323846;
const ll INF = 1e18;
const ld EPS = 1e-9;
const ll MAX_N = 202020;
const ll mod = 998244353;
typedef pair<ll, ll> pll;
typedef vector<pll> vpll;
typedef array<ll,3> all3;
typedef array<ll,5> all5;
typedef vector<all3> vall3;
typedef vector<all5> vall5;
typedef vector<ll> vll;
typedef vector<vll> vvll;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef unordered_set<ll> usll;
typedef unordered_set<pll, PairHash> uspll;
typedef unordered_map<ll, ll> umll;
#define rep(i,n) for(ll i=0;i<n;i++)
#define rrep(i,m,n) for(ll i=n;i>=m;i--)
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
ll __gcd(ll x, ll y) { return !y ? x : __gcd(y, x % y); }
ll __lcm(ll x, ll y) { return x * y / __gcd(x,y); }
using namespace std;
ll solve(ll n, ll m, vs& A) {
vvll vis(n,vll(m,INF));
vvll inc(n, vll(m));
vis[0][0] = 0;
queue<pll> q;
q.push({0,0});
ll dy[4]{-1,0,1,0}, dx[4]{0,1,0,-1};
while(!q.empty()) {
auto [y, x] = q.front(); q.pop();
inc[y][x] = false;
rep(i,4) {
ll ny = y + dy[i], nx = x + dx[i];
if(0 <= ny and ny < n and 0 <= nx and nx < m) {
if(A[ny][nx] == '.' and vis[ny][nx] > vis[y][x]) {
vis[ny][nx] = vis[y][x];
if(!inc[ny][nx]) {
q.push({ny, nx});
inc[ny][nx] = true;
}
}
}
}
for(ll ddy = -2; ddy <= 2; ddy++) {
for(ll ddx = -2; ddx <= 2; ddx++) {
if(abs(ddy) + abs(ddx) == 4) continue;
ll ny = y + ddy, nx = x + ddx;
if(0 <= ny and ny < n and 0 <= nx and nx < m) {
if(vis[ny][nx] > vis[y][x] + 1) {
vis[ny][nx] = vis[y][x] + 1;
if(!inc[ny][nx]) {
q.push({ny, nx});
inc[ny][nx] = true;
}
}
}
}
}
}
return vis[n-1][m-1];
}
int main() {
Code By Sumfi
ll tc = 1;
for (ll i = 1; i <= tc; i++) {
ll n,m;
cin>>n>>m;
vs A(n);
rep(i,n) cin>>A[i];
cout<<solve(n,m,A)<<endl;
}
return 0;
}
|
