結果

問題 No.2914 正閉路検出
コンテスト
ユーザー 👑 binap
提出日時 2024-10-04 23:03:30
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 3,675 bytes
コンパイル時間 4,454 ms
コンパイル使用メモリ 261,180 KB
最終ジャッジ日時 2025-02-24 15:35:27
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
other AC * 29 WA * 16 RE * 10
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 # 

#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<vector<tuple<int, int, long long>>> graph(n);
	vector<pair<int, int>> uv;
	rep(i, m){
		int u, v;
		long long c;
		cin >> u >> v >> c;
		u--; v--;
		graph[u].emplace_back(i + 1, v, c);
		graph[v].emplace_back(i + 1, u, -c);
		uv.emplace_back(u, v);
	}
	vector<int> check(n);
	vector<long long> potential(n);
	
	vector<int> ans;
	deque<int> de;
	int ans_from;
	
	auto dfs = [&](auto self, int from, long long prev) -> void{
		check[from] = 1;
		for(auto [idx, to, c] : graph[from]){
			if(ans.size()) return;
			if(idx == prev) continue;
			if(check[to]){
				if(potential[to] != potential[from] + c){
					int pos = from;
					ans.push_back(idx);
					while(pos != to){
						flush(cout);
						int idx = de.back(); de.pop_back();
						if(uv[idx].first == pos) pos = uv[idx].second;
						else pos = uv[idx].first;
						ans.push_back(idx);
					}
					ans_from = to;
					if(potential[to] > potential[from] + c){
					}else{
						reverse(ans.begin(), ans.end());
					}
				}
			}else{
				potential[to] = potential[from] + c;
				de.push_back(idx);
				self(self, to, idx);
				if(ans.size() == int(0)) de.pop_back();
			}
		}
	};
	
	rep(i, n){
		if(ans.size()) break;
		if(check[i]) continue;
		dfs(dfs, i, -1);
	}
	if(ans.size()){
		cout << ans.size() << "\n";
		cout << ans_from + 1 << "\n";
		cout << ans;
	}else{
		cout << "-1\n";
	}
	return 0;
}
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