結果

問題 No.2310 [Cherry 5th Tune A] Against Regret
ユーザー t98slidert98slider
提出日時 2023-05-19 23:30:07
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 695 ms / 6,000 ms
コード長 4,815 bytes
コンパイル時間 2,037 ms
コンパイル使用メモリ 185,416 KB
実行使用メモリ 6,820 KB
最終ジャッジ日時 2024-12-20 02:05:07
合計ジャッジ時間 14,823 ms
ジャッジサーバーID
(参考情報)
judge5 / judge4
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
other AC * 30
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>
using namespace std;
using ll = long long;
template<const unsigned int MOD> struct prime_modint {
using mint = prime_modint;
unsigned int v;
prime_modint() : v(0) {}
prime_modint(unsigned int a) { a %= MOD; v = a; }
prime_modint(unsigned long long a) { a %= MOD; v = a; }
prime_modint(int a) { a %= (int)(MOD); if(a < 0)a += MOD; v = a; }
prime_modint(long long a) { a %= (int)(MOD); if(a < 0)a += MOD; v = a; }
static constexpr int mod() { return MOD; }
mint& operator++() {v++; if(v == MOD)v = 0; return *this;}
mint& operator--() {if(v == 0)v = MOD; v--; return *this;}
mint operator++(int) { mint result = *this; ++*this; return result; }
mint operator--(int) { mint result = *this; --*this; return result; }
mint& operator+=(const mint& rhs) { v += rhs.v; if(v >= MOD) v -= MOD; return *this; }
mint& operator-=(const mint& rhs) { if(v < rhs.v) v += MOD; v -= rhs.v; return *this; }
mint& operator*=(const mint& rhs) {
v = (unsigned int)((unsigned long long)(v) * rhs.v % MOD);
return *this;
}
mint& operator/=(const mint& rhs) { return *this = *this * rhs.inv(); }
mint operator+() const { return *this; }
mint operator-() const { return mint() - *this; }
mint pow(long long n) const {
assert(0 <= n);
mint r = 1, x = *this;
while (n) {
if (n & 1) r *= x;
x *= x;
n >>= 1;
}
return r;
}
mint inv() const { assert(v); return pow(MOD - 2); }
friend mint operator+(const mint& lhs, const mint& rhs) { return mint(lhs) += rhs; }
friend mint operator-(const mint& lhs, const mint& rhs) { return mint(lhs) -= rhs; }
friend mint operator*(const mint& lhs, const mint& rhs) { return mint(lhs) *= rhs; }
friend mint operator/(const mint& lhs, const mint& rhs) { return mint(lhs) /= rhs; }
friend bool operator==(const mint& lhs, const mint& rhs) { return (lhs.v == rhs.v); }
friend bool operator!=(const mint& lhs, const mint& rhs) { return (lhs.v != rhs.v); }
friend std::ostream& operator << (std::ostream &os, const mint& rhs) noexcept { return os << rhs.v; }
};
//using mint = prime_modint<1000000007>;
using mint = prime_modint<998244353>;
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int n;
cin >> n;
vector<vector<mint>> A(n + 1, vector<mint>(n + 1));
vector<vector<mint>> B(n + 1, vector<mint>(n + 1));
vector<vector<mint>> dp(n + 1, vector<mint>(n + 1));
for(int i = 0; i <= n; i++){
for(int j = 0; j <= n; j++){
int v;
cin >> v;
A[i][j] = v;
}
}
for(int i = 0; i <= n; i++){
dp[i][i] = 1;
for(int j = i; j <= n; j++){
for(int k = j; k <= n; k++){
dp[i][k] += dp[i][j] * A[j][k];
}
}
}
/*int Q;
cin >> Q;
while(Q--){
int K, u, v, c;
cin >> K;
vector<tuple<int,int,int>> edge(K);
vector<int> b;
for(int i = 0; i < K; i++){
cin >> u >> v >> c;
edge[i] = make_tuple(u, v, c);
b.emplace_back(u);
b.emplace_back(v);
A[u][v] += c;
}
b.emplace_back(0);
b.emplace_back(n);
sort(b.begin(), b.end());
b.erase(unique(b.begin(), b.end()), b.end());
vector<mint> dp2(n + 1);
dp2[0] = 1;
for(int i = 0; i < n; i++){
for(int j = i + 1; j <= n; j++){
dp2[j] += dp2[i] * A[i][j];
}
}
cout << dp2.back() << '\n';
for(int i = 0; i < K; i++){
tie(u, v, c) = edge[i];
A[u][v] -= c;
}
}*/
int Q;
cin >> Q;
while(Q--){
int K, u, v, c;
cin >> K;
vector<tuple<int,int,int>> edge(K);
vector<int> b;
for(int i = 0; i < K; i++){
cin >> u >> v >> c;
edge[i] = make_tuple(u, v, c);
b.emplace_back(u);
b.emplace_back(v);
B[u][v] += c;
}
b.emplace_back(0);
b.emplace_back(n);
sort(b.begin(), b.end());
b.erase(unique(b.begin(), b.end()), b.end());
mint dp2[b.size()][2], ans = dp[0][n];
dp2[0][0] = 1;
for(int i = 0; i < b.size(); i++){
for(int j = i + 1; j < b.size(); j++){
dp2[j][1] += dp2[i][0] * dp[b[i]][b[j]];
dp2[j][0] += dp2[i][0] * B[b[i]][b[j]];
dp2[j][0] += dp2[i][1] * B[b[i]][b[j]];
}
}
cout << dp2[b.size() - 1][0] + dp2[b.size() - 1][1] << '\n';
for(int i = 0; i < K; i++){
tie(u, v, c) = edge[i];
B[u][v] -= c;
}
}
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0