結果
| 問題 | No.2411 Reverse Directions | 
| コンテスト | |
| ユーザー |  | 
| 提出日時 | 2023-08-11 21:54:57 | 
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) | 
| 結果 | 
                                WA
                                 
                             | 
| 実行時間 | - | 
| コード長 | 6,479 bytes | 
| コンパイル時間 | 2,051 ms | 
| コンパイル使用メモリ | 209,064 KB | 
| 最終ジャッジ日時 | 2025-02-16 01:20:03 | 
| ジャッジサーバーID (参考情報) | judge3 / judge4 | 
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| ファイルパターン | 結果 | 
|---|---|
| sample | AC * 3 | 
| other | AC * 18 WA * 11 | 
ソースコード
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (n); i++)
#define per(i, n) for (int i = (n)-1; i >= 0; i--)
#define rep2(i, l, r) for (int i = (l); i < (r); i++)
#define per2(i, l, r) for (int i = (r)-1; i >= (l); i--)
#define each(e, v) for (auto &e : v)
#define MM << " " <<
#define pb push_back
#define eb emplace_back
#define all(x) begin(x), end(x)
#define rall(x) rbegin(x), rend(x)
#define sz(x) (int)x.size()
using ll = long long;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
template <typename T>
using minheap = priority_queue<T, vector<T>, greater<T>>;
template <typename T>
using maxheap = priority_queue<T>;
template <typename T>
bool chmax(T &x, const T &y) {
    return (x < y) ? (x = y, true) : false;
}
template <typename T>
bool chmin(T &x, const T &y) {
    return (x > y) ? (x = y, true) : false;
}
template <typename T>
int flg(T x, int i) {
    return (x >> i) & 1;
}
int pct(int x) { return __builtin_popcount(x); }
int pct(ll x) { return __builtin_popcountll(x); }
int topbit(int x) { return (x == 0 ? -1 : 31 - __builtin_clz(x)); }
int topbit(ll x) { return (x == 0 ? -1 : 63 - __builtin_clzll(x)); }
int botbit(int x) { return (x == 0 ? -1 : __builtin_ctz(x)); }
int botbit(ll x) { return (x == 0 ? -1 : __builtin_ctzll(x)); }
template <typename T>
void print(const vector<T> &v, T x = 0) {
    int n = v.size();
    for (int i = 0; i < n; i++) cout << v[i] + x << (i == n - 1 ? '\n' : ' ');
    if (v.empty()) cout << '\n';
}
template <typename T>
void printn(const vector<T> &v, T x = 0) {
    int n = v.size();
    for (int i = 0; i < n; i++) cout << v[i] + x << '\n';
}
template <typename T>
int lb(const vector<T> &v, T x) {
    return lower_bound(begin(v), end(v), x) - begin(v);
}
template <typename T>
int ub(const vector<T> &v, T x) {
    return upper_bound(begin(v), end(v), x) - begin(v);
}
template <typename T>
void rearrange(vector<T> &v) {
    sort(begin(v), end(v));
    v.erase(unique(begin(v), end(v)), end(v));
}
template <typename T>
vector<int> id_sort(const vector<T> &v, bool greater = false) {
    int n = v.size();
    vector<int> ret(n);
    iota(begin(ret), end(ret), 0);
    sort(begin(ret), end(ret), [&](int i, int j) { return greater ? v[i] > v[j] : v[i] < v[j]; });
    return ret;
}
template <typename T>
void reorder(vector<T> &a, const vector<int> &ord) {
    int n = a.size();
    vector<T> b(n);
    for (int i = 0; i < n; i++) b[i] = a[ord[i]];
    swap(a, b);
}
template <typename T>
T floor(T x, T y) {
    assert(y != 0);
    if (y < 0) x = -x, y = -y;
    return (x >= 0 ? x / y : (x - y + 1) / y);
}
template <typename T>
T ceil(T x, T y) {
    assert(y != 0);
    if (y < 0) x = -x, y = -y;
    return (x >= 0 ? (x + y - 1) / y : x / y);
}
template <typename S, typename T>
pair<S, T> operator+(const pair<S, T> &p, const pair<S, T> &q) {
    return make_pair(p.first + q.first, p.second + q.second);
}
template <typename S, typename T>
pair<S, T> operator-(const pair<S, T> &p, const pair<S, T> &q) {
    return make_pair(p.first - q.first, p.second - q.second);
}
template <typename S, typename T>
istream &operator>>(istream &is, pair<S, T> &p) {
    S a;
    T b;
    is >> a >> b;
    p = make_pair(a, b);
    return is;
}
template <typename S, typename T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
    return os << p.first << ' ' << p.second;
}
struct io_setup {
    io_setup() {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout << fixed << setprecision(15);
        cerr << fixed << setprecision(15);
    }
} io_setup;
constexpr int inf = (1 << 30) - 1;
constexpr ll INF = (1LL << 60) - 1;
// constexpr int MOD = 1000000007;
constexpr int MOD = 998244353;
void solve() {
    int H, W, K, L, R;
    cin >> H >> W >> K >> L >> R;
    L--;
    vector<string> S(H);
    rep(i, H) cin >> S[i];
    if ((R - L) % 2 == 1) {
        cout << "No\n";
        return;
    }
    if ((H + W - 2) % 2 != K % 2) {
        cout << "No\n";
        return;
    }
    vector<int> dx = {1, 0, -1, 0}, dy = {0, 1, 0, -1};
    auto IN = [&](int i, int j) { return 0 <= i && i < H && 0 <= j && j < W && S[i][j] == '.'; };
    auto bfs = [&](int s, int t) {
        queue<pii> que;
        vector<vector<int>> d(H, vector<int>(W, inf));
        d[s][t] = 0;
        que.emplace(s, t);
        while (!empty(que)) {
            auto [x, y] = que.front();
            que.pop();
            rep(k, 4) {
                int nx = x + dx[k], ny = y + dy[k];
                if (IN(nx, ny)) {
                    if (chmin(d[nx][ny], d[x][y] + 1)) que.emplace(nx, ny);
                }
            }
        }
        return d;
    };
    auto d1 = bfs(0, 0), d2 = bfs(H - 1, W - 1);
    string T = "DRUL";
    rep(x, H) rep(y, W) {
        if (d1[x][y] <= L && d2[x][y] <= K - R) {
            string ans;
            {
                int px = x, py = y, r = d1[x][y];
                while (r > 0) {
                    rep(k, 4) {
                        int nx = px - dx[k], ny = py - dy[k];
                        if (IN(nx, ny) && d1[nx][ny] == r - 1) {
                            ans += T[k];
                            px = nx, py = ny;
                            r--;
                            break;
                        }
                    }
                }
                reverse(all(ans));
            }
            int c = K - d1[x][y] - d2[x][y];
            assert(c % 2 == 0);
            rep(k, 4) {
                if (IN(x + dx[k], y + dy[k])) {
                    rep(i, c / 2) {
                        ans += T[k];
                        ans += T[k ^ 2];
                    }
                    break;
                }
            }
            {
                int px = x, py = y, r = d2[x][y];
                while (r > 0) {
                    rep(k, 4) {
                        int nx = px + dx[k], ny = py + dy[k];
                        if (IN(nx, ny) && d2[nx][ny] == r - 1) {
                            ans += T[k];
                            px = nx, py = ny;
                            r--;
                            break;
                        }
                    }
                }
            }
            cout << "Yes\n";
            cout << ans << '\n';
            return;
        }
    }
    cout << "No\n";
}
int main() {
    int T = 1;
    // cin >> T;
    while (T--) solve();
}
            
            
            
        