結果
問題 | No.2733 Just K-times TSP |
ユーザー |
|
提出日時 | 2024-04-25 21:29:36 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 350 ms / 2,000 ms |
コード長 | 2,725 bytes |
コンパイル時間 | 4,410 ms |
コンパイル使用メモリ | 234,348 KB |
実行使用メモリ | 44,288 KB |
最終ジャッジ日時 | 2024-11-08 09:28:30 |
合計ジャッジ時間 | 7,415 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 2 |
other | AC * 32 |
ソースコード
#include<bits/stdc++.h>using namespace std;using ll = long long;using pll = pair<ll, ll>;#define drep(i, cc, n) for (ll i = (cc); i <= (n); ++i)#define rep(i, n) drep(i, 0, n - 1)#define all(a) (a).begin(), (a).end()#define pb push_back#define fi first#define se secondmt19937_64 rng(chrono::system_clock::now().time_since_epoch().count());const ll MOD1000000007 = 1000000007;const ll MOD998244353 = 998244353;const ll MOD[3] = {999727999, 1070777777, 1000000007};const ll LINF = 1LL << 60LL;const int IINF = (1 << 30) - 1;template<typename T>struct edge{int from, to;T cost;int id;edge(){}edge(int to, T cost=1) : from(-1), to(to), cost(cost){}edge(int to, T cost, int id) : from(-1), to(to), cost(cost), id(id){}edge(int from, int to, T cost, int id) : from(from), to(to), cost(cost), id(id){}};template<typename T>struct redge{int from, to;T cap, cost;int rev;redge(int to, T cap, T cost=(T)(1)) : from(-1), to(to), cap(cap), cost(cost){}redge(int to, T cap, T cost, int rev) : from(-1), to(to), cap(cap), cost(cost), rev(rev){}};template<typename T> using Edges = vector<edge<T>>;template<typename T> using weighted_graph = vector<Edges<T>>;template<typename T> using tree = vector<Edges<T>>;using unweighted_graph = vector<vector<int>>;template<typename T> using residual_graph = vector<vector<redge<T>>>;#include <atcoder/all>using namespace atcoder;using mint = modint998244353;void solve(){int n, m, k; cin >> n >> m >> k;unweighted_graph G(n);for(int i=0; i<m; i++){int u, v; cin >> u >> v;u--; v--;G[u].pb(v);G[v].pb(u);}vector<int> p(8, 1);for(int i=1; i<8; i++) p[i] = 10*p[i-1];vector<mint> dp(1e7, 0);vector<bool> visited(1e7, false);queue<int> Q;for(int v=0; v<n; v++){int state = p[v] + v*p[6];dp[state] = 1;visited[state] = true;Q.push(state);}while(!Q.empty()){int state = Q.front();Q.pop();int v = state/p[6];for(int to : G[v]){int cnt = (state/p[to])%10;if(cnt == k) continue;int nxt = to*p[6] + (state%p[6]) + p[to];if(!visited[nxt]){visited[nxt] = true;Q.push(nxt);}dp[nxt] += dp[state];}}mint ans = 0;int state = 0;for(int v=0; v<n; v++) state += p[v]*k;for(int v=0; v<n; v++){ans += dp[p[6]*v + state];}cout << ans.val() << '\n';}int main(){cin.tie(nullptr);ios::sync_with_stdio(false);int T=1;//cin >> T;while(T--) solve();}