题解 | #牛牛走迷宫#

逆序处理, 从前向后贪心, 每次选择字典序最小的字符加入到操作集合

#include <bits/stdc++.h>

#define x first
#define y second
#define all(x) x.begin(), x.end()
#define vec1(T, name, n, val) vector<T> name(n, val)
#define vec2(T, name, n, m, val) vector<vector<T>> name(n, vector<T>(m, val))
#define vec3(T, name, n, m, k, val) vector<vector<vector<T>>> name(n, vector<vector<T>>(m, vector<T>(k, val)))
#define vec4(T, name, n, m, k, p, val) vector<vector<vector<vector<T>>>> name((n), vector<vector<vector<T>>>((m), vector<vector<T>>((k), vector<T>((p), (val)))))

using namespace std;
using i128 = __int128;
using u128 = unsigned __int128;
using LL = long long;
using LD = long double;
using ULL = unsigned long long;
using PII = pair<int, int>;
using PLL = pair<LL, LL>;
using PLD = pair<LD, LD>;

const int N = 1e5 + 10, MOD = 998244353;
const int INF = 1e9;
const LL LL_INF = 2e18;
const LD EPS = 1e-8;
const int dx4[] = {1, 0, 0, -1}, dy4[] = {0, -1, 1, 0};
const int dx8[] = {-1, -1, -1, 0, 0, 1, 1, 1}, dy8[] = {-1, 0, 1, -1, 1, -1, 0, 1};
const int hx[] = {-2, -2, -1, -1, 1, 1, 2, 2}, hy[] = {-1, 1, -2, 2, -2, 2, -1, 1};

istream& operator>>(istream& is, i128& val) {
    string str;
    is >> str;
    val = 0;
    bool flag = false;
    if (str[0] == '-') flag = true, str = str.substr(1);
    for (char& c : str) val = val * 10 + c - '0';
    if (flag) val = -val;
    return is;
}

ostream& operator<<(ostream& os, i128 val) {
    if (val < 0) os << "-", val = -val;
    if (val > 9) os << val / 10;
    os << static_cast<char>(val % 10 + '0');
    return os;
}

bool cmp(LD a, LD b) {
    if (fabs(a - b) < EPS) return 1;
    return 0;
}

LL qpow(LL a, LL b) {
    LL ans = 1;
    a %= MOD;
    while (b) {
        if (b & 1) ans = ans * a % MOD;
        a = a * a % MOD;
        b >>= 1;
    }
    return ans;
}

void solve() {
    int n, m;
    cin >> n >> m;
    vector<string> g(n);
    for (int i = 0; i < n; ++i) cin >> g[i];

    vec2(int, d, n, m, -1);
    d[n - 1][m - 1] = 0;
    queue<PII> q;
    q.push({n - 1, m - 1});

    while (q.size()) {
        auto [x, y] = q.front();
        q.pop();
        for (int i = 0; i < 4; ++i) {
            int nx = x + dx4[i];
            int ny = y + dy4[i];
            if (nx < 0 || nx >= n || ny < 0 || ny >= m || g[nx][ny] == '1' || d[nx][ny] != -1) continue;
            d[nx][ny] = d[x][y] + 1;
            q.push({nx, ny});
        }
    }

    cout << d[0][0] << '\n';
    if (d[0][0] == -1) {
        return;
    }

    string ans;
    int x = 0, y = 0;
    while (x != n - 1 || y != m - 1) {
        for (int i = 0; i < 4; ++i) {
            int nx = x + dx4[i];
            int ny = y + dy4[i];
            if (nx < 0 || nx >= n || ny < 0 || ny >= m) continue;
            if (d[nx][ny] + 1 == d[x][y]) {
                if (i == 0) ans += 'D';
                else if (i == 1) ans += 'L';
                else if (i == 2) ans += 'R';
                else ans += 'U';
                x = nx, y = ny;
                break;
            }
        }
    }
    cout << ans << '\n';

/**/ #ifdef LOCAL
    cout << flush;
/**/ #endif
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    int T = 1;
    while (T--) solve();
    cout << fixed << setprecision(15);

    return 0;
}