結果

問題 No.34 砂漠の行商人
ユーザー fumiphys
提出日時 2020-04-10 21:10:45
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 549 ms / 5,000 ms
コード長 3,990 bytes
コンパイル時間 1,643 ms
コンパイル使用メモリ 172,472 KB
実行使用メモリ 17,536 KB
最終ジャッジ日時 2024-09-15 13:44:24
合計ジャッジ時間 4,958 ms
ジャッジサーバーID
（参考情報） 		judge2 / 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 * 26
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 # 

// includes
#include <bits/stdc++.h>
using namespace std;

// macros
#define pb emplace_back
#define mk make_pair
#define FOR(i, a, b) for(int i=(a);i<(b);++i)
#define rep(i, n) FOR(i, 0, n)
#define rrep(i, n) for(int i=((int)(n)-1);i>=0;i--)
#define irep(itr, st) for(auto itr = (st).begin(); itr != (st).end(); ++itr)
#define irrep(itr, st) for(auto itr = (st).rbegin(); itr != (st).rend(); ++itr)
#define whole(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define bit(n) (1LL<<(n))
// functions
template <typename T> void unique(T& c){c.erase(std::unique(c.begin(), c.end()), c.end());}
template <class T>bool chmax(T &a, const T &b){if(a < b){a = b; return 1;} return 0;}
template <class T>bool chmin(T &a, const T &b){if(a > b){a = b; return 1;} return 0;}
template <typename T> istream &operator>>(istream &is, vector<T> &vec){for(auto &v: vec)is >> v; return is;}
template <typename T> ostream &operator<<(ostream &os, const vector<T>& vec){for(int i = 0; i < vec.size(); i++){ os << vec[i]; if(i + 1 != vec.size())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_set<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T> ostream &operator<<(ostream &os, const unordered_multiset<T>& st){for(auto itr = st.begin(); itr != st.end(); ++itr){ os << *itr; auto titr = itr; if(++titr != st.end())os << " ";} return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p){os << "(" << p.first << ", " << p.second << ")"; return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
template <typename T1, typename T2> ostream &operator<<(ostream &os, const unordered_map<T1, T2> &mp){for(auto itr = mp.begin(); itr != mp.end(); ++itr){ os << "(" << itr->first << ", " << itr->second << ")"; auto titr = itr; if(++titr != mp.end())os << " "; } return os;}
//  types
using ll = long long int;
using P = pair<int, int>;
// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1000000007;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};
// io
struct fast_io{
  fast_io(){ios_base::sync_with_stdio(false); cin.tie(nullptr); cout << fixed << setprecision(20);}
} fast_io_;

int l[110][110];
const int b = 10001;
using bts = bitset<b>;
bts bt[10010];
int n;

int idx(int i, int j){
  return i * n + j;
}

struct State{
  int i, j, r, c;
  State(int i, int j, int r, int c): i(i), j(j), r(r), c(c){}
};

bool in(int i, int j){
  return (i >= 0) && (i < n) && (j >= 0) && (j < n);
}

int main(int argc, char const* argv[])
{
  int v; cin >> n >> v;
  int sx, sy, gx, gy; cin >> sx >> sy >> gx >> gy, sx--, sy--, gx--, gy--;
  swap(sx, sy);
  swap(gx, gy);
  rep(i, n){
    rep(j, n)cin >> l[i][j];
  }
  queue<State> q;
  q.push(State(sx, sy, v, 0));
  bt[idx(sx, sy)][v] = 1;
  while(!q.empty()){
    auto p = q.front(); q.pop();
    if(p.i == gx && p.j == gy){
      cout << p.c << endl;
      return 0;
    }
    rep(k, 4){
      int nx = p.i + dx[k], ny = p.j + dy[k];
      if(in(nx, ny) && p.r - l[nx][ny] > 0 && !bt[idx(nx, ny)][p.r - l[nx][ny]]){
        bt[idx(nx, ny)][p.r - l[nx][ny]] = 1;
        q.push(State(nx, ny, p.r - l[nx][ny], p.c + 1));
      }
    }
  }
  cout << -1 << endl;
  return 0;
}
0