結果

問題 No.34 砂漠の行商人
ユーザー fumiphys
提出日時 2019-08-20 22:07:06
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 1,159 ms / 5,000 ms
コード長 4,004 bytes
コンパイル時間 1,613 ms
コンパイル使用メモリ 175,072 KB
実行使用メモリ 61,028 KB
最終ジャッジ日時 2024-10-06 16:10:49
合計ジャッジ時間 7,766 ms
ジャッジサーバーID
（参考情報） 		judge3 / judge2
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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>

// macros
#define pb emplace_back
#define mk make_pair
#define pq priority_queue
#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 vrep(v, i) for(int i = 0; i < (v).size(); i++)
#define all(x) (x).begin(),(x).end()
#define sz(x) ((int)(x).size())
#define UNIQUE(v) v.erase(unique(v.begin(), v.end()), v.end())
#define FI first
#define SE second
#define bit(n) (1LL<<(n))
#define INT(n) int n; cin >> n;
#define LL(n) ll n; cin >> n;
#define DOUBLE(n) double n; cin >> n;
using namespace std;

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>;
using Pli = pair<ll, int>;
using Pil = pair<int, ll>;
using Pll = pair<ll, ll>;
using Pdd = pair<double, double>;
using cd = complex<double>;

// constants
const int inf = 1e9;
const ll linf = 1LL << 50;
const double EPS = 1e-10;
const int mod = 1e9 + 7;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, -1, 0, 1};

// solve
int l[110][110];
bool used[101][101][10001];

int main(int argc, char const* argv[])
{
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout << fixed << setprecision(20);
  INT(n); INT(v); INT(sx); INT(sy); INT(gx); INT(gy);
  sx--; sy--; gx--; gy--;
  rep(i, n){
    rep(j, n)cin >> l[j][i];
  }
  queue<pair<P, P>> q;
  q.push(mk(mk(sx, sy), mk(v, 0)));
  while(!q.empty()){
    auto p = q.front(); q.pop();
    int x = p.FI.FI;
    int y = p.FI.SE;
    int r = p.SE.FI;
    int d = p.SE.SE;
    if(used[x][y][r])continue;
    used[x][y][r] = true;
    if(x == gx && y == gy){
      cout << d << endl;
      return 0;
    }
    rep(k, 4){
      int nx = x + dx[k];
      int ny = y + dy[k];
      if(nx >= 0 && nx < n && ny >= 0 && ny < n && r > l[nx][ny]){
        if(!used[nx][ny][r-l[nx][ny]])q.push(mk(mk(nx, ny), mk(r - l[nx][ny], d + 1)));
      }
    }
  }
  cout << -1 << endl;
  return 0;
}
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