結果

問題 No.2411 Reverse Directions
ユーザー ecottea
提出日時 2023-08-11 22:27:45
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 23 ms / 2,000 ms
コード長 7,826 bytes
コンパイル時間 4,398 ms
コンパイル使用メモリ 263,912 KB
最終ジャッジ日時 2025-02-16 01:41:20
ジャッジサーバーID
(参考情報)
judge3 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 29
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ソースコード

diff #
プレゼンテーションモードにする

#ifndef HIDDEN_IN_VS //
//
#define _CRT_SECURE_NO_WARNINGS
//
#include <bits/stdc++.h>
using namespace std;
//
using ll = long long; // -2^63 2^63 = 9 * 10^18int -2^31 2^31 = 2 * 10^9
using pii = pair<int, int>; using pll = pair<ll, ll>; using pil = pair<int, ll>; using pli = pair<ll, int>;
using vi = vector<int>; using vvi = vector<vi>; using vvvi = vector<vvi>; using vvvvi = vector<vvvi>;
using vl = vector<ll>; using vvl = vector<vl>; using vvvl = vector<vvl>; using vvvvl = vector<vvvl>;
using vb = vector<bool>; using vvb = vector<vb>; using vvvb = vector<vvb>;
using vc = vector<char>; using vvc = vector<vc>; using vvvc = vector<vvc>;
using vd = vector<double>; using vvd = vector<vd>; using vvvd = vector<vvd>;
template <class T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
using Graph = vvi;
//
const double PI = acos(-1);
const vi DX = { 1, 0, -1, 0 }; // 4
const vi DY = { 0, 1, 0, -1 };
int INF = 1001001001; ll INFL = 4004004003104004004LL; // (int)INFL = 1010931620;
double EPS = 1e-15;
//
struct fast_io { fast_io() { cin.tie(nullptr); ios::sync_with_stdio(false); cout << fixed << setprecision(18); } } fastIOtmp;
//
#define all(a) (a).begin(), (a).end()
#define sz(x) ((int)(x).size())
#define lbpos(a, x) (int)distance((a).begin(), std::lower_bound(all(a), x))
#define ubpos(a, x) (int)distance((a).begin(), std::upper_bound(all(a), x))
#define Yes(b) {cout << ((b) ? "Yes\n" : "No\n");}
#define rep(i, n) for(int i = 0, i##_len = int(n); i < i##_len; ++i) // 0 n-1
#define repi(i, s, t) for(int i = int(s), i##_end = int(t); i <= i##_end; ++i) // s t
#define repir(i, s, t) for(int i = int(s), i##_end = int(t); i >= i##_end; --i) // s t
#define repe(v, a) for(const auto& v : (a)) // a
#define repea(v, a) for(auto& v : (a)) // a
#define repb(set, d) for(int set = 0; set < (1 << int(d)); ++set) // d
#define repp(a) sort(all(a)); for(bool a##_perm = true; a##_perm; a##_perm = next_permutation(all(a))) // a
#define smod(n, m) ((((n) % (m)) + (m)) % (m)) // mod
#define uniq(a) {sort(all(a)); (a).erase(unique(all(a)), (a).end());} //
#define EXIT(a) {cout << (a) << endl; exit(0);} //
#define inQ(x, y, u, l, d, r) ((u) <= (x) && (l) <= (y) && (x) < (d) && (y) < (r)) //
//
template <class T> inline ll pow(T n, int k) { ll v = 1; rep(i, k) v *= n; return v; }
template <class T> inline bool chmax(T& M, const T& x) { if (M < x) { M = x; return true; } return false; } // true
    
template <class T> inline bool chmin(T& m, const T& x) { if (m > x) { m = x; return true; } return false; } // true
    
template <class T> inline T get(T set, int i) { return (set >> i) & T(1); }
//
template <class T, class U> inline istream& operator>>(istream& is, pair<T, U>& p) { is >> p.first >> p.second; return is; }
template <class T> inline istream& operator>>(istream& is, vector<T>& v) { repea(x, v) is >> x; return is; }
template <class T> inline vector<T>& operator--(vector<T>& v) { repea(x, v) --x; return v; }
template <class T> inline vector<T>& operator++(vector<T>& v) { repea(x, v) ++x; return v; }
#endif //
#if __has_include(<atcoder/all>)
#include <atcoder/all>
using namespace atcoder;
#ifdef _MSC_VER
#include "localACL.hpp"
#endif
//using mint = modint1000000007;
using mint = modint998244353;
//using mint = modint; // mint::set_mod(m);
namespace atcoder {
inline istream& operator>>(istream& is, mint& x) { ll x_; is >> x_; x = x_; return is; }
inline ostream& operator<<(ostream& os, const mint& x) { os << x.val(); return os; }
}
using vm = vector<mint>; using vvm = vector<vm>; using vvvm = vector<vvm>; using vvvvm = vector<vvvm>;
#endif
#ifdef _MSC_VER // Visual Studio
#include "local.hpp"
#else // gcc
inline int popcount(int n) { return __builtin_popcount(n); }
inline int popcount(ll n) { return __builtin_popcountll(n); }
inline int lsb(int n) { return n != 0 ? __builtin_ctz(n) : -1; }
inline int lsb(ll n) { return n != 0 ? __builtin_ctzll(n) : -1; }
inline int msb(int n) { return n != 0 ? (31 - __builtin_clz(n)) : -1; }
inline int msb(ll n) { return n != 0 ? (63 - __builtin_clzll(n)) : -1; }
#define gcd __gcd
#define dump(...)
#define dumpel(v)
#define dump_list(v)
#define dump_mat(v)
#define input_from_file(f)
#define output_to_file(f)
#define Assert(b) { if (!(b)) while (1) cout << "OLE"; }
#endif
//O(h w)
/*
* wall h×w c s = (sx, sy)
* c[i][j] dist[i][j] INF
*
*
*/
void solve_maze(const vvc& c, const pii& s, vvi& dist, const char wall = '#') {
// verify : https://algo-method.com/tasks/424
int h = sz(c), w = sz(c[0]);
dist = vvi(h, vi(w, INF));
dist[s.first][s.second] = 0;
// q :
queue<pii> q;
q.push(s);
while (!q.empty()) {
int x, y;
tie(x, y) = q.front(); q.pop();
// (x, y) 4 調
rep(k, 4) {
// (nx, ny) : (x, y)
int nx = x + DX[k];
int ny = y + DY[k];
//
if (nx < 0 || nx >= h || ny < 0 || ny >= w || c[nx][ny] == wall) continue;
//
if (dist[nx][ny] != INF) continue;
//
dist[nx][ny] = dist[x][y] + 1;
q.push({ nx, ny });
}
}
}
int main() {
input_from_file("input.txt");
// output_to_file("output.txt");
int h, w, k, l, r;
cin >> h >> w >> k >> l >> r;
l--;
vvc s(h, vc(w));
cin >> s;
if (((h - 1) + (w - 1)) % 2 != k % 2) EXIT("No");
if ((r - l) % 2 == 1) EXIT("No");
vvi safe(h, vi(w));
repi(i, 1, h - 2) repi(j, 0, w - 1) {
safe[i][j] ^= (s[i - 1][j] == '.' && s[i][j] == '.' && s[i + 1][j] == '.');
}
repi(i, 0, h - 1) repi(j, 1, w - 2) {
safe[i][j] ^= 2 * (s[i][j - 1] == '.' && s[i][j] == '.' && s[i][j + 1] == '.');
}
rep(i, h) rep(j, w) if ((i + j) % 2 != l % 2) safe[i][j] = 0;
dumpel(safe);
vvi distS, distT;
solve_maze(s, { 0, 0 }, distS);
solve_maze(s, { h - 1, w - 1 }, distT);
dumpel(distS); dumpel(distT);
string DRUL = "DRUL";
rep(i, h) rep(j, w) {
if (safe[i][j] == 0) continue;
if (distS[i][j] > l || distT[i][j] > (k - r)) continue;
if (distS[i][j] + (r - l) + distT[i][j] > k) continue;
string res1, res2, res3;
int x = i, y = j; int d = distS[i][j];
while (d > 0) {
rep(k, 4) {
int px = x + DX[k];
int py = y + DY[k];
if (inQ(px, py, 0, 0, h, w) && distS[px][py] == d - 1) {
res1.push_back(DRUL[k ^ 2]);
x = px;
y = py;
d--;
break;
}
}
}
reverse(all(res1));
dump(res1);
rep(hoge, (k - distS[i][j] - distT[i][j]) / 2) {
if (safe[i][j] & 1) res2 += "UD";
else res2 += "RL";
}
dump(res2);
x = i; y = j; d = distT[i][j];
while (d > 0) {
rep(k, 4) {
int px = x + DX[k];
int py = y + DY[k];
if (inQ(px, py, 0, 0, h, w) && distT[px][py] == d - 1) {
res3.push_back(DRUL[k]);
x = px;
y = py;
d--;
break;
}
}
}
dump(res3);
Yes(1);
cout << res1 << res2 << res3 << endl;
return 0;
}
EXIT("No");
}
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