結果

問題 No.3079 Unite Japanese Prefectures
ユーザー 👑 binap
提出日時 2025-02-19 01:12:37
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,308 ms / 4,000 ms
コード長 3,199 bytes
コンパイル時間 4,200 ms
コンパイル使用メモリ 268,792 KB
実行使用メモリ 56,448 KB
最終ジャッジ日時 2025-03-16 10:22:36
合計ジャッジ時間 13,927 ms
ジャッジサーバーID
（参考情報） 		judge4 / judge5
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 * 27
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;
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);}
int main(){
    int n, m;
    cin >> n >> m;
    
    vector<tuple<int, int, int>> G;
    
    rep(i, m){
        int u, v, c;
        cin >> u >> v >> c;
        u--; v--; c--;
        G.emplace_back(c, u, v);
    }
    sort(G.begin(), G.end());
    
    atcoder::dsu uf(n);
    vector<int> cnt(6);
    for(auto [c, u, v] : G){
        if(uf.same(u, v)) continue;
        cnt[c]++;
        uf.merge(u, v);
    }
    
    map<vector<int>, long double> memo;
    vector<int> zero = {0, 0, 0, 0, 0, 0};
    auto solve = [&](vector<int> cond, auto solve) -> long double{
        if(cond == zero) return (long double)0;
        if(memo.find(cond) != memo.end()) return memo[cond];
        int MAX = -1;
        int ng = 0;
        long double sum = 0;
        for(int i = 0; i <= 5; i++){
            if(cond[i] > 0) MAX = i;
            if(MAX >= 0){
                cond[MAX]--;
                sum += solve(cond, solve);
                cond[MAX]++;
            }else ng++;
        }
        assert(ng < 6);
        return memo[cond] = (sum + 6) / (6 - ng);
    };
    cout << setprecision(15);
    cout << solve(cnt, solve) << "\n";
    return 0;
}
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