#900999 (C++17) No.2411 Reverse Directions

提出ソース

結果

問題	No.2411 Reverse Directions
ユーザー	dyktr_06
提出日時	2023-08-11 23:18:49
言語	C++17 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	30 ms / 2,000 ms
コード長	7,568 bytes
コンパイル時間	7,058 ms
コンパイル使用メモリ	321,328 KB
実行使用メモリ	9,052 KB
最終ジャッジ日時	2024-11-18 18:41:43
合計ジャッジ時間	8,734 ms
ジャッジサーバーID （参考情報）	judge3 / judge2

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	2 ms 6,820 KB
testcase_01	AC	2 ms 6,816 KB
testcase_02	AC	2 ms 6,816 KB
testcase_03	AC	2 ms 6,816 KB
testcase_04	AC	6 ms 6,816 KB
testcase_05	AC	2 ms 6,820 KB
testcase_06	AC	2 ms 6,820 KB
testcase_07	AC	5 ms 6,816 KB
testcase_08	AC	2 ms 6,816 KB
testcase_09	AC	1 ms 6,816 KB
testcase_10	AC	15 ms 7,308 KB
testcase_11	AC	2 ms 6,820 KB
testcase_12	AC	30 ms 8,616 KB
testcase_13	AC	2 ms 6,816 KB
testcase_14	AC	8 ms 6,820 KB
testcase_15	AC	3 ms 6,820 KB
testcase_16	AC	4 ms 6,820 KB
testcase_17	AC	11 ms 6,820 KB
testcase_18	AC	4 ms 6,816 KB
testcase_19	AC	2 ms 6,820 KB
testcase_20	AC	8 ms 6,816 KB
testcase_21	AC	2 ms 6,816 KB
testcase_22	AC	19 ms 7,280 KB
testcase_23	AC	2 ms 6,816 KB
testcase_24	AC	28 ms 9,052 KB
testcase_25	AC	2 ms 6,816 KB
testcase_26	AC	2 ms 6,820 KB
testcase_27	AC	2 ms 6,816 KB
testcase_28	AC	13 ms 6,820 KB
testcase_29	AC	26 ms 8,284 KB
testcase_30	AC	17 ms 6,820 KB
testcase_31	AC	2 ms 6,820 KB

権限があれば一括ダウンロードができます

ソースコード

raw source code

#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 << ")"; return os; }
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();
}

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結果

テストケース

ソースコード