結果
問題 | No.2411 Reverse Directions |
ユーザー |
|
提出日時 | 2023-08-11 23:18:49 |
言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 26 ms / 2,000 ms |
コード長 | 7,568 bytes |
コンパイル時間 | 23,377 ms |
コンパイル使用メモリ | 359,284 KB |
最終ジャッジ日時 | 2025-02-16 02:07:42 |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 29 |
ソースコード
#pragma GCC target("avx2")#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#include <bits/stdc++.h>#include <atcoder/all>using namespace std;using namespace atcoder;#define overload4(_1, _2, _3, _4, name, ...) name#define rep1(n) for(int i = 0; i < (int)(n); ++i)#define rep2(i, n) for(int i = 0; i < (int)(n); ++i)#define rep3(i, a, b) for(int i = (a); i < (int)(b); ++i)#define rep4(i, a, b, c) for(int i = (a); i < (int)(b); i += (c))#define rep(...) overload4(__VA_ARGS__, rep4, rep3, rep2, rep1)(__VA_ARGS__)#define rrep(i,n) for(int i = (int)(n) - 1; i >= 0; --i)#define ALL(a) a.begin(), a.end()#define Sort(a) sort(a.begin(), a.end())#define RSort(a) sort(a.rbegin(), a.rend())typedef long long int ll;typedef unsigned long long ul;typedef long double ld;typedef vector<int> vi;typedef vector<long long> vll;typedef vector<char> vc;typedef vector<string> vst;typedef vector<double> vd;typedef vector<long double> vld;typedef pair<long long, long long> P;template<class T> long long sum(const T& a){ return accumulate(a.begin(), a.end(), 0LL); }template<class T> auto min(const T& a){ return *min_element(a.begin(), a.end()); }template<class T> auto max(const T& a){ return *max_element(a.begin(), a.end()); }const long long MINF = 0x7fffffffffff;const long long INF = 0x1fffffffffffffff;const long long MOD = 998244353;const long double EPS = 1e-9;const long double PI = acos(-1);template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p){ is >> p.first >> p.second; return is; }template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){ os << "(" << p.first << ", " << p.second << ")"; returnos; }template<typename T> istream &operator>>(istream &is, vector<T> &v){ for(T &in : v) is >> in; return is; }template<typename T> ostream &operator<<(ostream &os, const vector<T> &v){ for(int i = 0; i < (int) v.size(); ++i){ os << v[i] << (i + 1 != (int) v.size() ? " " : ""); } return os; }template <typename T, typename S> ostream &operator<<(ostream &os, const map<T, S> &mp){ for(auto &[key, val] : mp){ os << key << ":" << val << " ";} return os; }template <typename T> ostream &operator<<(ostream &os, const set<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os << *itr<< (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }template <typename T> ostream &operator<<(ostream &os, const multiset<T> &st){ auto itr = st.begin(); for(int i = 0; i < (int)st.size(); ++i){ os <<*itr << (i + 1 != (int)st.size() ? " " : ""); itr++; } return os; }template <typename T> ostream &operator<<(ostream &os, queue<T> q){ while(q.size()){ os << q.front() << " "; q.pop(); } return os; }template <typename T> ostream &operator<<(ostream &os, deque<T> q){ while(q.size()){ os << q.front() << " "; q.pop_front(); } return os; }template <typename T> ostream &operator<<(ostream &os, stack<T> st){ while(st.size()){ os << st.top() << " "; st.pop(); } return os; }template <class T, class Container, class Compare> ostream &operator<<(ostream &os, priority_queue<T, Container, Compare> pq){ while(pq.size()){ os<< pq.top() << " "; pq.pop(); } return os; }template<class T, class U> inline T vin(T& vec, U n) { vec.resize(n); for(int i = 0; i < (int) n; ++i) cin >> vec[i]; return vec; }template<class T> inline void vout(T vec, string s = "\n"){ for(auto x : vec) cout << x << s; }template<class... T> void in(T&... a){ (cin >> ... >> a); }void out(){ cout << '\n'; }template<class T, class... Ts> void out(const T& a, const Ts&... b){ cout << a; (cout << ... << (cout << ' ', b)); cout << '\n'; }template<class T, class U> void inGraph(vector<vector<T>>& G, U n, U m, bool directed = false){ G.resize(n); for(int i = 0; i < m; ++i){ int a, b;cin >> a >> b; a--, b--; G[a].push_back(b); if(!directed) G[b].push_back(a); } }ll h, w, k, l, r;vst s;void input(){in(h, w, k, l, r);vin(s, h);}vector<int> dx = {0, 1, 0, -1};vector<int> dy = {1, 0, -1, 0};vector<vector<int>> bfs(const vector<string> &s, vector<pair<int, int>> sxy = {{0, 0}}){int h = s.size(), w = s[0].size();deque<tuple<int, int, int>> dq;for(auto [sx, sy] : sxy){dq.push_back(make_tuple(0, sx, sy));}vector<vector<int>> d(h, vector<int>(w, -1));while (!dq.empty()){int dd = get<0>(dq.front());int x = get<1>(dq.front());int y = get<2>(dq.front());dq.pop_front();if(d[x][y] == -1){d[x][y] = dd;for(int i = 0; i < 4; i++){int x2 = x + dx[i];int y2 = y + dy[i];if(0 <= x2 && x2 < h && 0 <= y2 && y2 < w){if(s[x2][y2] == '.'){// 優先的に見るかどうかで前に入れるか後ろに入れるかを決めるdq.push_back(make_tuple(dd + 1, x2, y2));}}}}}return d;}vc dirc = {'L', 'U', 'R', 'D'};void solve(){if((r - l + 1) % 2 != 0){out("No");return;}if((h + w) % 2 != k % 2){out("No");return;}vector<vi> cost = bfs(s);vector<vc> memo(h, vc(w, '$'));vector<pair<int, int>> sxy;rep(i, h) rep(j, w){if(cost[i][j] % 2 == (l - 1) % 2 && cost[i][j] <= l - 1 && cost[i][j] != -1){vll dircnt(2);rep(dir, 4){ll x2 = i + dx[dir];ll y2 = j + dy[dir];if(0 <= x2 && x2 < h && 0 <= y2 && y2 < w){if(s[x2][y2] == '.'){dircnt[dir % 2]++;}}}if(dircnt[0] == 2){memo[i][j] = 'R';}else if(dircnt[1] == 2){memo[i][j] = 'D';}if(memo[i][j] != '$'){sxy.emplace_back(i, j);}}}vector<vi> cost2 = bfs(s, sxy);if(cost2[h - 1][w - 1] > k - r || cost2[h - 1][w - 1] == -1){out("No");return;}string ans = "";ll nowx = h - 1, nowy = w - 1;while(cost2[nowx][nowy] != 0){rep(dir, 4){ll x2 = nowx + dx[dir];ll y2 = nowy + dy[dir];if(0 <= x2 && x2 < h && 0 <= y2 && y2 < w){if(cost2[x2][y2] == cost2[nowx][nowy] - 1){ans += dirc[dir];nowx = x2, nowy = y2;break;}}}}ll cnt = k - cost[nowx][nowy] - cost2[h - 1][w - 1];rep(i, cnt / 2){if(memo[nowx][nowy] == 'R'){ans += "RL";}else{ans += "UD";}}while(cost[nowx][nowy] != 0){rep(dir, 4){ll x2 = nowx + dx[dir];ll y2 = nowy + dy[dir];if(0 <= x2 && x2 < h && 0 <= y2 && y2 < w){if(cost[x2][y2] == cost[nowx][nowy] - 1){ans += dirc[dir];nowx = x2, nowy = y2;break;}}}}reverse(ALL(ans));out("Yes");out(ans);}int main(){ios::sync_with_stdio(false);cin.tie(nullptr);cout << fixed << setprecision(20);input();solve();}