結果

問題 No.2733 Just K-times TSP
ユーザー 👑 binap
提出日時 2024-04-20 04:50:58
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,724 ms / 2,000 ms
コード長 2,264 bytes
コンパイル時間 4,420 ms
コンパイル使用メモリ 256,296 KB
最終ジャッジ日時 2025-02-21 06:02:52
ジャッジサーバーID
(参考情報)
judge1 / judge4
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 * 2
other AC * 32
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 vector<int> vi;
typedef vector<long long> vl;
typedef vector<vector<int>> vvi;
typedef vector<vector<long long>> vvl;
typedef long double ld;
typedef pair<int, int> P;
ostream& operator<<(ostream& os, const modint& a) {os << a.val(); return os;}
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<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> 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 n, m, k;
cin >> n >> m >> k;
vector<vector<int>> G(n, vector<int>(n));
rep(i, m){
int u, v;
cin >> u >> v;
u--; v--;
G[u][v] = 1;
G[v][u] = 1;
}
auto f = [&](vector<int> cnt){
int pos = 0;
rep(i, n){
pos *= (k + 1);
pos += cnt[i];
}
return pos;
};
auto f_inv = [&](int pos){
vector<int> cnt(n);
rep(i, n){
cnt[(n - 1) - i] = pos % (k + 1);
pos /= k + 1;
}
return cnt;
};
int x = 1;
rep(i, n) x *= k + 1;
vector<vector<mint>> dp(x, vector<mint>(n));
rep(i, n){
vector<int> cnt(n);
cnt[i] = 1;
int pos = f(cnt);
dp[pos][i] = 1;
}
rep(pos_from, x){
rep(from, n){
rep(to, n){
if(G[from][to] == 1){
vector<int> cnt = f_inv(pos_from);
if(cnt[to] >= k) continue;
cnt[to]++;
dp[f(cnt)][to] += dp[pos_from][from];
}
}
}
}
mint ans = 0;
rep(i, n) ans += dp[x - 1][i];
cout << ans << "\n";
return 0;
}
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