結果

問題 No.2916 累進コスト最小化
ユーザー 👑 binap
提出日時 2024-05-25 20:12:07
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 310 ms / 3,500 ms
コード長 2,976 bytes
コンパイル時間 4,528 ms
コンパイル使用メモリ 259,224 KB
最終ジャッジ日時 2025-02-21 16:39:13
ジャッジサーバーID
（参考情報） 		judge2 / judge4
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 * 34
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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;
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};
    }
    for(int c = 1; c <= C; c++){
        vector<int> D(N, -1);
        D[0] = c;
        set<pair<int, int>> E;
        rep(i, N) E.insert({D[i], i});
        while(E.size()){
            auto it = prev(E.end());
            auto [money_from, from] = *it;
            E.erase(it);
            if(money_from == -1) continue;
            vector<int> cand;
            for(auto [tmp, to] : E) cand.push_back(to);
            for(int to : cand){
                auto [r, w] = G[from][to];
                if(r == -1) continue;
                int money_to = money_from - money_from / r - w;
                if(money_to > D[to]){
                    E.erase({D[to], to});
                    D[to] = money_to;
                    E.insert({D[to], to});
                }
            }
        }
        cout << D[N - 1] << "\n";
    }
    return 0;
}
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