結果

問題 No.177 制作進行の宮森あおいです！
ユーザー Haar
提出日時 2019-03-28 23:30:17
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 3 ms / 2,000 ms
コード長 3,665 bytes
コンパイル時間 2,129 ms
コンパイル使用メモリ 187,340 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-15 00:35:36
合計ジャッジ時間 2,570 ms
ジャッジサーバーID
（参考情報） 		judge3 / 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
sample AC * 3
other AC * 13
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>
#define FOR(v, a, b) for(int v = (a); v < (b); ++v)
#define FORE(v, a, b) for(int 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(int v = (a); v >= (b); --v)
#define ALL(x) (x).begin(), (x).end()
#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 LLI long long int
#define fst first
#define snd second
#ifdef DEBUG
#include <misc/C++/Debug.cpp>
#else
#define dump(x) ((void)0)
#endif
using namespace std;
#define gcd __gcd
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> istream& operator>>(istream &is, pair<T,U> &p){is >> p.first >> p.second; return is;}
template <typename T, typename U> T& chmin(T &a, const U &b){return a = (a<=b?a:b);}
template <typename T, typename U> T& chmax(T &a, const U &b){return a = (a>=b?a:b);}
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, T INF> class Dinic{
private:
  vector<vector<pair<int,T>>> graph;
  int size, s, t;
  vector<vector<T>> cap;
  vector<int> level;
  
  bool buildLevel(){
    fill(ALL(level), 0);
    level[s] = 1;
    deque<int> deq = {s};
    while(!deq.empty()){
      int cur = deq.front(); deq.pop_front();
      REP(i,size)
    if(level[i]==0 && cap[cur][i]>0){
      level[i] = level[cur] + 1;
      deq.push_back(i);
    }
    }
    return level[t] != 0;
  }
  void dfs(vector<int> &path, T &flow){
    if(path.empty()) return;
    int cur = path.back();
    if(cur == t){
      T f = INF;
      FOR(i,1,(int)path.size()) f = min(f, cap[path[i-1]][path[i]]);
      FOR(i,1,(int)path.size()){
    cap[path[i-1]][path[i]] -= f;
    cap[path[i]][path[i-1]] += f;
      }
      flow += f;
    }else{
      REP(i,size){
    if(cap[cur][i]>0 && level[i]>level[cur]){
      path.push_back(i);
      dfs(path, flow);
      path.pop_back();
    }
      }
    }
  }
  T augment(){
    T f = 0;
    vector<int> path = {s};
    dfs(path, f);
    return f;
  }
  T loop(){
    T f = 0;
    while(buildLevel()) f += augment();
    return f;
  }
public:
  Dinic(vector<vector<pair<int,T>>> &_graph): graph(_graph), size(graph.size()) {}
  Dinic(int size): graph(size), size(size){}
  void add_edge(int from, int to, const T &cap){
    graph[from].push_back({to, cap});
  }
  
  T max_flow(int _s, int _t){
    cap = vector<vector<T>>(size, vector<T>(size, 0));
    level = vector<int>(size, 0);
    REP(i,size)
      for(auto &p : graph[i]){
    int j = p.first;
    T d = p.second;
    cap[i][j] += d;
      }
    s = _s;
    t = _t;
    return loop();
  }
};
int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int W;
  while(cin >> W){
    int N; cin >> N;
    vector<int> J(N); cin >> J;
    int M; cin >> M;
    vector<int> C(M); cin >> C;
    Dinic<int,INT_MAX> mflow(N+M+2);
    int s = N+M, t = s+1;
    REP(i,N) mflow.add_edge(s, i, J[i]);
    REP(i,M) mflow.add_edge(N+i, t, C[i]);
    REP(i,M){
      int Q; cin >> Q;
      vector<int> X(Q); cin >> X;
      vector<bool> t(N,true);
      REP(j,Q) t[X[j]-1] = false;
      REP(j,N){
    if(t[j]) mflow.add_edge(j, N+i, INT_MAX);
      }
    }
    int res = mflow.max_flow(s, t);
    cout << (res<W ? "BANSAKUTSUKITA" : "SHIROBAKO") << endl;
  }
  
  return 0;
}
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