結果

問題 No.2916 累進コスト最小化
コンテスト
ユーザー 👑 binap
提出日時 2024-05-26 10:59:52
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 75 ms / 3,500 ms
コード長 2,910 bytes
コンパイル時間 4,322 ms
コンパイル使用メモリ 256,812 KB
最終ジャッジ日時 2025-02-21 16:43:39
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
（要ログイン）
ファイルパターン 結果
other AC * 34
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;

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> 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);}

using P = pair<int, int>;

int main(){
	int N, M, C;
	cin >> N >> M >> C;
	vector<vector<P>> G(N, vector<P>(N, {-1, -1}));
	rep(_, M){
		int u, v, r, w;
		cin >> u >> v >> r >> w;
		u--; v--;
		G[u][v] = {r, w};
		G[v][u] = {r, w};
	}
	int ans = -1;
	vector<int> best(N, -1);
	for(int c = 1; c <= C; c++){
		queue<pair<int, int>> qu;
		qu.emplace(0, c);
		while(qu.size()){
			auto [from, c_from] = qu.front(); qu.pop();
			if(c_from <= best[from]) continue;
			best[from] = c_from;
			if(from == N - 1) ans = c_from;
			for(int to = 0; to < N; to++){
				auto [r, w] = G[from][to];
				if(r == -1) continue;
				int c_to = c_from - (c_from / r + w);
				if(c_to < 0) continue;
				if(c_to <= best[to]) continue;
				qu.emplace(to, c_to);
			}
		}
		cout << ans << "\n";
	}
	return 0;
}
0