#369412 (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	2019-08-16 22:12:11
言語	C++14 (gcc 12.3.0 + boost 1.83.0)
結果	AC
実行時間	30 ms / 5,000 ms
コード長	5,388 bytes
コンパイル時間	2,149 ms
コンパイル使用メモリ	183,712 KB
実行使用メモリ	7,936 KB
最終ジャッジ日時	2024-09-22 17:11:32
合計ジャッジ時間	2,941 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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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	2 ms 6,812 KB
testcase_01	AC	2 ms 6,940 KB
testcase_02	AC	2 ms 6,940 KB
testcase_03	AC	3 ms 6,940 KB
testcase_04	AC	3 ms 6,944 KB
testcase_05	AC	23 ms 7,468 KB
testcase_06	AC	30 ms 7,680 KB
testcase_07	AC	29 ms 7,808 KB
testcase_08	AC	21 ms 7,808 KB
testcase_09	AC	29 ms 7,936 KB
testcase_10	AC	2 ms 6,944 KB
testcase_11	AC	2 ms 6,940 KB
testcase_12	AC	3 ms 6,940 KB
testcase_13	AC	3 ms 6,944 KB
testcase_14	AC	4 ms 6,944 KB
testcase_15	AC	3 ms 6,940 KB
testcase_16	AC	6 ms 6,944 KB
testcase_17	AC	5 ms 6,944 KB
testcase_18	AC	6 ms 6,940 KB
testcase_19	AC	7 ms 6,940 KB
testcase_20	AC	2 ms 6,940 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--; swap(ox, oy);
  rep(i, n)rep(j, n)cin >> l[i][j];
  GraphL g(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){
          g.adde(i * n + j, nx * n + ny, l[nx][ny]);
        }
      }
    }
  }
  g.dijkstra(0);
  bool ok = false;
  if(g.dis[n * n - 1] < v)ok = true;
  if(ox >= 0 && oy >= 0){
    if(g.dis[ox * n + oy] < v){
      ll tmp = 2 * (v - g.dis[ox * n + oy]);
      g.dijkstra(ox * n + oy);
      if(tmp > g.dis[n * n - 1])ok = true;
    }
  }
  if(ok)cout << "YES" << endl;
  else cout << "NO" << endl;
  return 0;
}

yukicoder

結果

テストケース

ソースコード