結果

問題 No.2411 Reverse Directions
ユーザー ktr216ktr216
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
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 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 -
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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

#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) 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";
}
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