結果

問題 No.2733 Just K-times TSP
ユーザー だれ
提出日時 2024-04-19 23:03:25
言語 C++23
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 274 ms / 2,000 ms
コード長 6,302 bytes
コンパイル時間 4,040 ms
コンパイル使用メモリ 230,996 KB
実行使用メモリ 19,072 KB
最終ジャッジ日時 2024-10-11 17:11:13
合計ジャッジ時間 6,242 ms
ジャッジサーバーID
(参考情報)
judge1 / judge2
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
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 <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <complex>
#include <cstdio>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <sstream>
#include <stack>
#include <unordered_set>
using namespace std;
#if __has_include(<atcoder/all>)
#include <atcoder/all>
#endif
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define _rep(i, n) _rep2(i, 0, n)
#define _rep2(i, a, b) for (int i = (int)(a); i < (int)(b); i++)
#define rep(...) GET_MACRO(__VA_ARGS__, _rep2, _rep)(__VA_ARGS__)
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define UNIQUE(x) \
std::sort((x).begin(), (x).end()); \
(x).erase(std::unique((x).begin(), (x).end()), (x).end())
using i64 = long long;
template <class T, class U>
bool chmin(T& a, const U& b) {
return (b < a) ? (a = b, true) : false;
}
template <class T, class U>
bool chmax(T& a, const U& b) {
return (b > a) ? (a = b, true) : false;
}
template <class T = std::string, class U = std::string>
inline void YesNo(bool f = 0, const T yes = "Yes", const U no = "No") {
if (f)
std::cout << yes << "\n";
else
std::cout << no << "\n";
}
namespace io {
template <class T, class U>
istream& operator>>(istream& i, pair<T, U>& p) {
i >> p.first >> p.second;
return i;
}
template <typename T>
istream& operator>>(istream& i, vector<T>& v) {
rep(j, v.size()) i >> v[j];
return i;
}
template <typename T>
string join(vector<T>& v) {
stringstream s;
rep(i, v.size()) s << ' ' << v[i];
return s.str().substr(1);
}
template <typename T>
ostream& operator<<(ostream& o, vector<T>& v) {
if (v.size()) o << join(v);
return o;
}
template <typename T>
string join(vector<vector<T>>& vv) {
string s = "\n";
rep(i, vv.size()) s += join(vv[i]) + "\n";
return s;
}
template <typename T>
ostream& operator<<(ostream& o, vector<vector<T>>& vv) {
if (vv.size()) o << join(vv);
return o;
}
template <class T, class U>
ostream& operator<<(ostream& o, pair<T, U>& p) {
o << p.first << " " << p.second;
return o;
}
void OUT() { std::cout << "\n"; }
template <class Head, class... Tail>
void OUT(Head&& head, Tail&&... tail) {
std::cout << head;
if (sizeof...(tail)) std::cout << ' ';
OUT(std::forward<Tail>(tail)...);
}
void OUTL() { std::cout << std::endl; }
template <class Head, class... Tail>
void OUTL(Head&& head, Tail&&... tail) {
std::cout << head;
if (sizeof...(tail)) std::cout << ' ';
OUTL(std::forward<Tail>(tail)...);
}
void IN() {}
template <class Head, class... Tail>
void IN(Head&& head, Tail&&... tail) {
cin >> head;
IN(std::forward<Tail>(tail)...);
}
} // namespace io
using namespace io;
namespace useful {
long long modpow(long long a, long long b, long long mod) {
long long res = 1;
while (b) {
if (b & 1) res *= a, res %= mod;
a *= a;
a %= mod;
b >>= 1;
}
return res;
}
bool is_pow2(long long x) { return x > 0 && (x & (x - 1)) == 0; }
template <class T>
void rearrange(vector<T>& a, vector<int>& p) {
vector<T> b = a;
for (int i = 0; i < int(a.size()); i++) {
a[i] = b[p[i]];
}
return;
}
template <class T>
vector<pair<T, int>> rle_sequence(vector<T>& a) {
vector<pair<T, int>> res;
int n = a.size();
int l = 1;
rep(i, n - 1) {
if (a[i] == a[i + 1])
l++;
else {
res.emplace_back(a[i], l);
l = 1;
}
}
res.emplace_back(a.back(), l);
return res;
}
vector<pair<char, int>> rle_string(string a) {
vector<pair<char, int>> res;
int n = a.size();
if (n == 1) return vector<pair<char, int>>{{a[0], 1}};
int l = 1;
rep(i, n - 1) {
if (a[i] == a[i + 1])
l++;
else {
res.emplace_back(a[i], l);
l = 1;
}
}
res.emplace_back(a.back(), l);
return res;
}
vector<int> linear_sieve(int n) {
vector<int> primes;
vector<int> res(n + 1);
iota(all(res), 0);
for (int i = 2; i <= n; i++) {
if (res[i] == i) primes.emplace_back(i);
for (auto j : primes) {
if (j * i > n) break;
res[j * i] = j;
}
}
return res;
// return primes;
}
template <class T>
vector<long long> dijkstra(vector<vector<pair<int, T>>>& graph, int start) {
int n = graph.size();
vector<long long> res(n, 2e18);
res[start] = 0;
priority_queue<pair<long long, int>, vector<pair<long long, int>>,
greater<pair<long long, int>>>
que;
que.push({0, start});
while (!que.empty()) {
auto [c, v] = que.top();
que.pop();
if (res[v] < c) continue;
for (auto [nxt, cost] : graph[v]) {
auto x = c + cost;
if (x < res[nxt]) {
res[nxt] = x;
que.push({x, nxt});
}
}
}
return res;
}
} // namespace useful
using namespace useful;
using mint = atcoder::modint998244353;
bool visited[9][9][9][9][9][9][6];
mint memo[9][9][9][9][9][9][6];
int main() {
cin.tie(nullptr);
ios::sync_with_stdio(false);
i64 n, m, k;
IN(n, m, k);
vector mat(n, vector<int>(n));
rep(i, m) {
int a, b;
IN(a, b);
a--, b--;
mat[a][b] = mat[b][a] = 1;
}
vector<int> target(6);
rep(i, n) target[i] = k;
auto rec = [&](auto&& self, vector<int>& v, int now) -> mint {
if (v == target) return 1;
if (visited[v[0]][v[1]][v[2]][v[3]][v[4]][v[5]][now]) {
return memo[v[0]][v[1]][v[2]][v[3]][v[4]][v[5]][now];
}
visited[v[0]][v[1]][v[2]][v[3]][v[4]][v[5]][now] = 1;
mint res = 0;
rep(i, n) {
if (mat[now][i] && v[i] < k) {
v[i]++;
res += self(self, v, i);
v[i]--;
}
}
return memo[v[0]][v[1]][v[2]][v[3]][v[4]][v[5]][now] = res;
};
vector<int> tmp(6);
mint ans = 0;
rep(i, n) {
tmp[i] = 1;
ans += rec(rec, tmp, i);
tmp[i] = 0;
}
OUT(ans.val());
}
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