#363896 (C++14) No.20 砂漠のオアシス

提出ソース

結果

問題	No.20 砂漠のオアシス
ユーザー	fumiphys
提出日時	2019-07-30 00:36:21
言語	C++14 (gcc 12.3.0 + boost 1.83.0)
結果	WA
実行時間	-
コード長	5,851 bytes
コンパイル時間	1,627 ms
コンパイル使用メモリ	185,888 KB
実行使用メモリ	8,448 KB
最終ジャッジ日時	2024-07-02 15:30:45
合計ジャッジ時間	2,475 ms
ジャッジサーバーID （参考情報）	judge3 / judge5

このコードへのチャレンジ
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テストケース

テストケース表示

入力	結果	実行時間実行使用メモリ
testcase_00	AC	1 ms 5,248 KB
testcase_01	AC	1 ms 5,376 KB
testcase_02	AC	1 ms 5,376 KB
testcase_03	AC	3 ms 5,376 KB
testcase_04	AC	3 ms 5,376 KB
testcase_05	AC	20 ms 7,596 KB
testcase_06	WA	-
testcase_07	AC	24 ms 8,448 KB
testcase_08	AC	24 ms 8,448 KB
testcase_09	WA	-
testcase_10	WA	-
testcase_11	AC	1 ms 5,376 KB
testcase_12	WA	-
testcase_13	WA	-
testcase_14	AC	4 ms 5,376 KB
testcase_15	AC	3 ms 5,376 KB
testcase_16	AC	6 ms 5,376 KB
testcase_17	WA	-
testcase_18	AC	5 ms 5,376 KB
testcase_19	WA	-
testcase_20	AC	2 ms 5,376 KB

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ソースコード

raw source code

// 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
template <typename T>
struct Graph {
  int n;
  vector<vector<pair<int, T> > > edge;
  vector<T> dis;
  Graph(int n): n(n) {
    edge.resize(n);
    dis.resize(n);
  }
  void dijkstra(int s){
    dijkstra(s, 0);
  }
  T dijkstra(int s, int t){
    // initialize
    fill(dis.begin(), dis.end(), -1);
    vector<bool> used(n, false);
    dis[s] = 0;
    // dijkstra
    priority_queue<pair<T, int>, vector<pair<T, int> >, greater<pair<T, int> > > q;
    q.push(make_pair(0, s));
    while(!q.empty()){
      pair<T, int> p = q.top(); q.pop();
      int at = p.second;
      T distance = p.first;
      if(used[at])continue;
      used[at] = true;
      for(auto itr = edge[at].begin(); itr != edge[at].end(); ++itr){
        int to = (*itr).first;
        T cost = (*itr).second;
        if(used[to])continue;
        if(dis[to] == -1 || dis[to] > distance + cost){
          q.push(make_pair(distance + cost, to));
          dis[to] = distance + cost;
        }
      }
    }
    return dis[t];
  }
  void adde(int at, int to, T cost){
    edge[at].push_back(make_pair(to, cost));
  }
  [[deprecated("This function takes O(|edge[at]|).")]]
  void remove(int at, int to){
    int index = -1;
    for(int i = 0; i < edge[at].size(); i++){
      if(edge[at][i].first == to){
        index = i;
        break;
      }
    }
    edge[at].erase(edge[at].begin() + index);
  }
};

using GraphI = Graph<int>;
using GraphL = Graph<ll>;
using GraphD = Graph<double>;

int l[210][210];

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(ox); INT(oy); ox--; oy--;
  rep(i, n)rep(j, n)cin >> l[i][j];
  GraphL graph(n * n);
  rep(i, n){
    rep(j, n){
      rep(k, 4){
        int nx = i + dx[k];
        int ny = j + dy[k];
        if(nx >= 0 && nx < n && ny >= 0 && ny < n){
          graph.adde(i * n + j, nx * n + ny, l[nx][ny]);
        }
      }
    }
  }
  graph.dijkstra(0);
  vector<ll> diso = graph.dis;
  if(ox >= 0 && oy >= 0){
    graph.dijkstra(ox * n + oy);
    vector<ll> disx = graph.dis;
    rep(k, 4){
      int nx = ox + dx[k];
      int ny = oy + dy[k];
      ll tmp = diso[ox * n + oy];
      if(nx >= 0 && nx < n && ny >= 0 && ny < n){
        if(tmp < v){
          ll rem = 2 * (v - tmp);
          rem -= l[nx][ny] - l[ox][oy];
          if(rem > v - tmp){
            cout << "YES" << endl;
            return 0;
          }
        }
      }
    }
    ll tmp = diso[ox * n + oy];
    if(tmp < v){
      ll rem = 2 * (v - tmp);
      if(rem > disx[n * n - 1]){
        cout << "YES" << endl;
        return 0;
      }
    }
  }
  if(diso[n * n - 1] < v){
    cout << "YES" << endl;
    return 0;
  }
  cout << "NO" << endl;
  return 0;
}

yukicoder

結果

テストケース

ソースコード