結果

問題 No.3071 Double Speedrun
ユーザー 👑 binap
提出日時 2025-03-21 23:15:49
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 593 ms / 6,000 ms
コード長 3,222 bytes
コンパイル時間 5,098 ms
コンパイル使用メモリ 257,316 KB
実行使用メモリ 7,324 KB
最終ジャッジ日時 2025-03-21 23:16:00
合計ジャッジ時間 10,178 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge3
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 14
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

diff #
プレゼンテーションモードにする

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
using namespace atcoder;
typedef long long ll;
typedef pair<int, int> P;
template <int m> ostream& operator<<(ostream& os, const static_modint<m>& a) {os << a.val(); return os;}
template <int m> ostream& operator<<(ostream& os, const dynamic_modint<m>& a) {os << a.val(); return os;}
template <int m> istream& operator>>(istream& is, static_modint<m>& a) {long long x; is >> x; a = x; return is;}
template <int m> istream& operator>>(istream& is, dynamic_modint<m>& a) {long long x; is >> x; a = x; return is;}
template<typename T> istream& operator>>(istream& is, vector<T>& v){int n = v.size(); assert(n > 0); rep(i, n) is >> v[i]; return is;}
template<typename U, typename T> ostream& operator<<(ostream& os, const pair<U, T>& p){os << p.first << ' ' << p.second; return os;}
template<typename T> ostream& operator<<(ostream& os, const vector<T>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : " "); return 
    os;}
template<typename T> ostream& operator<<(ostream& os, const vector<vector<T>>& v){int n = v.size(); rep(i, n) os << v[i] << (i == n - 1 ? "\n" : ""); 
    return os;}
template<typename T> ostream& operator<<(ostream& os, const set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename T> ostream& operator<<(ostream& os, const unordered_set<T>& se){for(T x : se) os << x << " "; os << "\n"; return os;}
template<typename S, auto op, auto e> ostream& operator<<(ostream& os, const atcoder::segtree<S, op, e>& seg){int n = seg.max_right(0, [](S){return 
    true;}); rep(i, n) os << seg.get(i) << (i == n - 1 ? "\n" : " "); return os;}
template<typename S, auto op, auto e, typename F, auto mapping, auto composition, auto id> ostream& operator<<(ostream& os, const atcoder
    ::lazy_segtree<S, op, e, F, mapping, composition, id>& seg){int n = seg.max_right(0, [](S){return true;}); rep(i, n) os << seg.get(i) << (i == n - 
    1 ? "\n" : " "); return os;}
template<typename T> void chmin(T& a, T b){a = min(a, b);}
template<typename T> void chmax(T& a, T b){a = max(a, b);}
using mint = modint998244353;
int main(){
    int h, w;
    cin >> h >> w;
    vector<string> s(h);
    cin >> s;
    
    vector<vector<mint>> dp(h + 1, vector<mint>(h + 1));
    for(int y = 0; y < h; y++){
        if(s[y][0] == '#') break;
        dp[0][y] = 1;
    }
    auto get_next = [&](int y, int x){
        for(int z = y; z < h; z++) if(s[z][x] == '#') return z;
        return h;
    };
    for(int x = 1; x < w; x++){
        vector<vector<mint>> dp_old(h + 1, vector<mint>(h + 1));
        swap(dp, dp_old);
        int na = get_next(0, x);
        for(int a = 0; a < h; a++){
            if(a == na){
                na = get_next(a + 1, x);
                continue;
            }
            int nb = get_next(a, x);
            for(int b = a + 1; b < h; b++){
                if(b == nb){
                    nb = get_next(b + 1, x);
                    continue;
                }
                mint val = dp_old[a][b];
                dp[a][b] += val;
                dp[min(na, b)][b] -= val;
                dp[a][nb] -= val;
                dp[min(na, b)][nb] += val;
            }
        }
        for(int a = 0; a < h; a++){
            for(int b = 0; b < h; b++){
                dp[a][b + 1] += dp[a][b];
            }
        }
        for(int a = 0; a < h; a++){
            for(int b = 0; b < h; b++){
                dp[a + 1][b] += dp[a][b];
            }
        }
    }
    cout << dp[h - 2][h - 1] << "\n";
    
    return 0;
}
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