結果
問題 | No.2411 Reverse Directions |
ユーザー | ktr216 |
提出日時 | 2023-08-11 22:27:03 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
MLE
|
実行時間 | - |
コード長 | 6,908 bytes |
コンパイル時間 | 2,448 ms |
コンパイル使用メモリ | 190,124 KB |
実行使用メモリ | 1,629,100 KB |
最終ジャッジ日時 | 2024-11-18 17:06:14 |
合計ジャッジ時間 | 58,624 ms |
ジャッジサーバーID (参考情報) |
judge2 / judge4 |
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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 | MLE | - |
testcase_01 | AC | 9 ms
19,676 KB |
testcase_02 | MLE | - |
testcase_03 | MLE | - |
testcase_04 | MLE | - |
testcase_05 | AC | 10 ms
20,936 KB |
testcase_06 | MLE | - |
testcase_07 | AC | 12 ms
20,008 KB |
testcase_08 | MLE | - |
testcase_09 | AC | 10 ms
19,356 KB |
testcase_10 | MLE | - |
testcase_11 | WA | - |
testcase_12 | MLE | - |
testcase_13 | AC | 15 ms
19,304 KB |
testcase_14 | MLE | - |
testcase_15 | MLE | - |
testcase_16 | MLE | - |
testcase_17 | MLE | - |
testcase_18 | MLE | - |
testcase_19 | AC | 26 ms
12,780 KB |
testcase_20 | MLE | - |
testcase_21 | TLE | - |
testcase_22 | MLE | - |
testcase_23 | AC | 16 ms
12,860 KB |
testcase_24 | TLE | - |
testcase_25 | AC | 11 ms
14,084 KB |
testcase_26 | TLE | - |
testcase_27 | TLE | - |
testcase_28 | TLE | - |
testcase_29 | TLE | - |
testcase_30 | TLE | - |
testcase_31 | MLE | - |
ソースコード
#include <bits/stdc++.h>using namespace std;#define double long doubleusing 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) 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) 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, R - L + 1) {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";}