結果

問題 No.2411 Reverse Directions
ユーザー ktr216ktr216
提出日時 2023-08-11 22:37:31
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 6,956 bytes
コンパイル時間 2,077 ms
コンパイル使用メモリ 181,304 KB
実行使用メモリ 6,456 KB
最終ジャッジ日時 2024-04-29 13:43:19
合計ジャッジ時間 5,667 ms
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 AC 3 ms
5,376 KB
testcase_01 AC 3 ms
5,376 KB
testcase_02 AC 3 ms
5,376 KB
testcase_03 AC 3 ms
5,376 KB
testcase_04 AC 8 ms
6,192 KB
testcase_05 AC 4 ms
5,376 KB
testcase_06 AC 3 ms
5,376 KB
testcase_07 AC 4 ms
5,632 KB
testcase_08 AC 3 ms
5,376 KB
testcase_09 AC 3 ms
5,376 KB
testcase_10 WA -
testcase_11 WA -
testcase_12 WA -
testcase_13 AC 4 ms
5,376 KB
testcase_14 WA -
testcase_15 WA -
testcase_16 AC 4 ms
5,376 KB
testcase_17 WA -
testcase_18 WA -
testcase_19 AC 4 ms
5,376 KB
testcase_20 WA -
testcase_21 AC 11 ms
5,504 KB
testcase_22 WA -
testcase_23 AC 4 ms
5,504 KB
testcase_24 WA -
testcase_25 AC 3 ms
5,376 KB
testcase_26 AC 5 ms
5,376 KB
testcase_27 AC 9 ms
5,376 KB
testcase_28 WA -
testcase_29 WA -
testcase_30 WA -
testcase_31 AC 23 ms
5,504 KB
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ソースコード

diff # 

#include <bits/stdc++.h>
using namespace std;

#define double long double

using ll = long long;
using VB = vector<bool>;
using VVB = vector<VB>;
using VVVB = vector<VVB>;
using VC = vector<char>;
using VVC = vector<VC>;
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 VD = vector<double>;
using VVD = vector<VD>;
using VVVD = vector<VVD>;
//using P = pair<int, int>;

#define REP(i, n) for (ll i = 0; i < (int)(n); i++)
#define FOR(i, a, b) for (ll i = a; i < (ll)(b); i++)
#define ALL(a) (a).begin(),(a).end()

constexpr int INF = 1001001001;
constexpr ll LINF = 1001001001001001001ll;
constexpr int DX[] = {1, 0, -1, 0};
constexpr int DY[] = {0, 1, 0, -1};

template< typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {return a < b && (a = b, true); }
template< typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {return a > b && (a = b, true); }

const ll MOD = 998244353;

/*

const int MAX_N = 200000;
int par[MAX_N];
int rnk[MAX_N];
int siz[MAX_N];

void init(int n) {
    REP(i,n) {
        par[i] = i;
        rnk[i] = 0;
        siz[i] = 1;
    }
}

int find(int x) {
    if (par[x] == x) {
        return x;
    } else {
        return par[x] = find(par[x]);
    }
}

void unite(int x, int y) {
    x = find(x);
    y = find(y);
    if (x == y) return;
    int s = siz[x] + siz[y];
    if (rnk[x] < rnk[y]) {
        par[x] = y;
    } else {
        par[y] = x;
        if (rnk[x] == rnk[y]) rnk[x]++;
    }
    siz[find(x)] = s;
}

bool same(int x, int y) {
    return find(x) == find(y);
}

int size(int x) {
    return siz[find(x)];
}

*/

ll mod_pow(ll x, ll n, ll mod) {
    ll res = 1;
    x %= mod;
    while (n > 0) {
        if (n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

ll gcd(ll x, ll y) {
    if (y == 0) return x;
    return gcd(y, x % y);
}

/*

typedef pair<ll, int> P0;
struct edge { int to; ll cost; };

const int MAX_V = 100000;
//const ll LINF = 1LL<<60;

int V;
vector<edge> G[MAX_V];
ll d[MAX_V];

void dijkstra(ll s) {
    priority_queue<P0, vector<P0>, greater<P0> > que;
    fill(d, d + V, LINF);
    d[s] = 0;
    que.push(P0(0, s));

    while (!que.empty()) {
        P0 p = que.top(); que.pop();
        int v = p.second;
        if (d[v] < p.first) continue;
        for (edge e : G[v]) {
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P0(d[e.to], e.to));
            }
        }
    }
}

*/

/*

double EPS = 1e-10;

double add(double a, double b) {
    if (abs(a + b) < EPS * (abs(a) + abs(b))) return 0;
    return a + b;
}

struct P {
    double x, y;
    P() {}
    P(double x, double y) : x(x), y(y) {
    }
    P operator + (P p) {
        return P(add(x, p.x), add(y, p.y));
    }
    P operator - (P p) {
        return P(add(x, -p.x), add(y, -p.y));
    }
    P operator * (double d) {
        return P(x * d, y * d);
    }
    double dot(P p) {
        return add(x * p.x, y * p.y);
    }
    double det(P p) {
        return add(x * p.y, -y * p.x);
    }
};

bool on_seg(P p1, P p2, P q) {
    return ()
}

P intersection(P p1, P p2, P q1, P q2) {
    return p1 + (p2 - p1) * ((q2 - q1).det(q1 - p1) / (q2 - q1).det(p2 - p1));
}

*/

/*

VL f(600010, 1);

ll C(ll n, ll k) {
    return f[n] * mod_pow(f[k], MOD - 2, MOD) % MOD * mod_pow(f[n - k], MOD - 2, MOD) % MOD;
}

ll P(ll n, ll k) {
    return f[n] * mod_pow(f[n - k], MOD - 2, MOD) % MOD;
}

*/

int H, W, K, L, R;
vector<string> S(500);
VVI D1(500, VI(500, INF)), D2(500, VI(500, INF));

int main() {
    ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    //REP(i, 600009) f[i + 1] = f[i] * (i + 1) % MOD;
    cin >> H >> W >> K >> L >> R;
    REP(i, H) cin >> S[i];
    if (L % 2 == R % 2) {
        cout << "No\n";
        return 0;
    }
    queue<VI> q;
    q.push({0, 0, 0});
    D1[0][0] = 0;
    while (!q.empty()) {
        VI v = q.front();
        q.pop();
        int x = v[0], y = v[1], d = v[2];
        REP(i, 4) {
            int nx = x + DX[i], ny = y + DY[i];
            if (nx < 0 || nx >= H || ny < 0 || ny >= W) continue;
            if (S[nx][ny] == '#') continue;
            if (D1[nx][ny] <= d + 1) continue;
            q.push({nx, ny, d + 1});
            D1[nx][ny] = d + 1;
        }
    }
    q.push({H - 1, W - 1, 0});
    D2[H - 1][W - 1] = 0;
    while (!q.empty()) {
        VI v = q.front();
        q.pop();
        int x = v[0], y = v[1], d = v[2];
        REP(i, 4) {
            int nx = x + DX[i], ny = y + DY[i];
            if (nx < 0 || nx >= H || ny < 0 || ny >= W) continue;
            if (S[nx][ny] == '#') continue;
            if (D2[nx][ny] <= d + 1) continue;
            q.push({nx, ny, d + 1});
            D2[nx][ny] = d + 1;
        }
    }
    /*
    REP(i, H) {
        REP(j, W) cout << D1[i][j] << "-" << D2[i][j] << " ";
        cout << endl;
    }
    */
    REP(i, H) REP(j, W) {
        if (D1[i][j] > L - 1) continue;
        if (D2[i][j] > K - R) continue;
        if ((D1[i][j] + L + 1) % 2) continue;
        if ((D2[i][j] + K + R) % 2) continue;
        if (S[i][j] == '#') continue;
        if ((i == 0 || i == H - 1 || S[i - 1][j] == '#' || S[i + 1][j] == '#') && (j == 0 || j == W - 1 || S[i][j - 1] == '#' || S[i][j + 1] == '#')) continue;
        string ans = "", T = "RDLU";
        int x = i, y = j, d = D1[i][j];
        while (d) {
            REP(k, 4) {
                int nx = x + DX[k], ny = y + DY[k];
                if (nx < 0 || nx >= H || ny < 0 || ny >= W) continue;
                if (D1[nx][ny] == d - 1) {
                    ans.push_back(T[(k + 2) % 4]);
                    d--;
                    x = nx;
                    y = ny;
                    break;
                }
            }
            //cout << "d=" << d << " ans1=" << ans1 << endl;
        }        
        reverse(ALL(ans));
        REP(k, K - D1[i][j] - D2[i][j]) {
            if (i > 0 && i < H - 1 && S[i - 1][j] == '.' && S[i + 1][j] == ',') {
                if (k % 2) ans.push_back('U');
                else ans.push_back('D');
            } else {
                if (k % 2) ans.push_back('L');
                else ans.push_back('R');
            }
        }
        x = i;
        y = j;
        d = D2[i][j];
        while (d) {
            REP(k, 4) {
                int nx = x + DX[k], ny = y + DY[k];
                if (nx < 0 || nx >= H || ny < 0 || ny >= W) continue;
                if (D2[nx][ny] == d - 1) {
                    ans.push_back(T[k ]);
                    d--;
                    x = nx;
                    y = ny;
                    break;
                }
            }
            //cout << "d=" << d << " ans1=" << ans1 << endl;
        }        
        cout << "Yes\n" << ans << endl;
        return 0;
    }
    cout << "No\n";
}
0