結果

問題 No.17 2つの地点に泊まりたい
コンテスト
ユーザー Haar
提出日時 2020-01-23 06:50:48
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 4 ms / 5,000 ms
コード長 4,069 bytes
コンパイル時間 2,714 ms
コンパイル使用メモリ 206,032 KB
最終ジャッジ日時 2025-01-08 20:32:40
ジャッジサーバーID
（参考情報） 		judge5 / judge1
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 * 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 # 

#include <bits/stdc++.h>
#define LLI long long int
#define FOR(v, a, b) for(LLI v = (a); v < (b); ++v)
#define FORE(v, a, b) for(LLI v = (a); v <= (b); ++v)
#define REP(v, n) FOR(v, 0, n)
#define REPE(v, n) FORE(v, 0, n)
#define REV(v, a, b) for(LLI v = (a); v >= (b); --v)
#define ALL(x) (x).begin(), (x).end()
#define RALL(x) (x).rbegin(), (x).rend()
#define ITR(it, c) for(auto it = (c).begin(); it != (c).end(); ++it)
#define RITR(it, c) for(auto it = (c).rbegin(); it != (c).rend(); ++it)
#define EXIST(c,x) ((c).find(x) != (c).end())
#define fst first
#define snd second
#define popcount __builtin_popcount
#define UNIQ(v) (v).erase(unique(ALL(v)), (v).end())
#define bit(i) (1LL<<(i))

#ifdef DEBUG
#include <misc/C++/Debug.cpp>
#else
#define dump(...) ((void)0)
#endif

#define gcd __gcd

using namespace std;
template <class T> constexpr T lcm(T m, T n){return m/gcd(m,n)*n;}

template <typename I> void join(ostream &ost, I s, I t, string d=" "){for(auto i=s; i!=t; ++i){if(i!=s)ost<<d; ost<<*i;}ost<<endl;}
template <typename T> istream& operator>>(istream &is, vector<T> &v){for(auto &a : v) is >> a; return is;}
template <typename T, typename U> bool chmin(T &a, const U &b){return (a>b ? a=b, true : false);}
template <typename T, typename U> bool chmax(T &a, const U &b){return (a<b ? a=b, true : false);}
template <typename T, size_t N, typename U> void fill_array(T (&a)[N], const U &v){fill((U*)a, (U*)(a+N), v);}
template <typename T> auto make_vector(int n, int m, const T &value){return vector<vector<T>>(n, vector<T>(m, value));}


struct Init{
  Init(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << fixed << setprecision(12);
    cerr << fixed << setprecision(12);
  }
}init;



template <typename Cost = int> class Edge{
public:
  int from,to;
  Cost cost;
  Edge() {}
  Edge(int to, Cost cost): to(to), cost(cost){}
  Edge(int from, int to, Cost cost): from(from), to(to), cost(cost){}

  Edge rev() const {return Edge(to,from,cost);}
  
  friend std::ostream& operator<<(std::ostream &os, const Edge &e){
    os << "(FROM: " << e.from << "," << "TO: " << e.to << "," << "COST: " << e.cost << ")";
    return os;
  }
};

template <typename T> using Graph = std::vector<std::vector<Edge<T>>>;
template <typename T> using Tree = std::vector<std::vector<Edge<T>>>;

template <typename C, typename T> void add_edge(C &g, int from, int to, T w){
  g[from].push_back(Edge<T>(from, to, w));  
}

template <typename C, typename T> void add_undirected(C &g, int a, int b, T w){
  g[a].push_back(Edge<T>(a, b, w));
  g[b].push_back(Edge<T>(b, a, w));
}

template <typename T>
struct WarshallFloyd{
  const int n;
  std::vector<std::vector<std::optional<T>>> dist;
  bool has_negative_cycle;
  
  WarshallFloyd(const Graph<T> &graph):
    n(graph.size()),
    dist(n, std::vector<std::optional<T>>(n, std::nullopt)),
    has_negative_cycle(false)
  {
    for(int i = 0; i < n; ++i) dist[i][i] = 0;
    
    for(int i = 0; i < n; ++i){
      for(auto &e : graph[i]){
        dist[e.from][e.to] = e.cost;
      }
    }

    for(int k = 0; k < n; ++k){
      for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
          if(dist[i][k] and dist[k][j]){
            if(not dist[i][j]){
              dist[i][j] = *dist[i][k] + *dist[k][j];
            }else{
              dist[i][j] = std::min(*dist[i][j], *dist[i][k] + *dist[k][j]);
            }
          }
        }
      }
    }
    
    for(int i = 0; i < n; ++i) if(*dist[i][i] < 0) has_negative_cycle = true;
  }
};




int main(){
  int N;
  while(cin >> N){
    Graph<int> g(N);

    vector<int> S(N); cin >> S;

    int M; cin >> M;
    REP(i,M){
      int a,b,c; cin >> a >> b >> c;
      add_undirected(g, a, b, c);
    }

    auto dist = WarshallFloyd(g).dist;

    int ans = INT_MAX;

    FOR(i,1,N-1){
      FOR(j,1,N-1){
        if(i == j) continue;
        
        if(dist[0][i] and dist[i][j] and dist[j][N-1]){
          chmin(ans, *dist[0][i] + *dist[i][j] + *dist[j][N-1] + S[i] + S[j]);
        }
      }
    }

    cout << ans << endl;
  }

  return 0;
}
0