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 00:30:53 |
言語 | C++14 (gcc 12.3.0 + boost 1.83.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 5,959 bytes |
コンパイル時間 | 1,664 ms |
コンパイル使用メモリ | 185,604 KB |
実行使用メモリ | 8,576 KB |
最終ジャッジ日時 | 2024-07-02 15:30:48 |
合計ジャッジ時間 | 2,425 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge1 |
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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
5,248 KB |
testcase_01 | AC | 2 ms
5,376 KB |
testcase_02 | AC | 2 ms
5,376 KB |
testcase_03 | AC | 2 ms
5,376 KB |
testcase_04 | AC | 2 ms
5,376 KB |
testcase_05 | AC | 22 ms
7,596 KB |
testcase_06 | WA | - |
testcase_07 | AC | 24 ms
8,448 KB |
testcase_08 | AC | 25 ms
8,448 KB |
testcase_09 | WA | - |
testcase_10 | AC | 2 ms
5,376 KB |
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 |
ソースコード
// 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){ if(nx == ox && ny == oy)graph.adde(i * n + j, nx * n + ny, inf); else 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]; if(nx >= 0 && nx < n && ny >= 0 && ny < n){ ll tmp = diso[nx * n + ny]; tmp += l[ox][oy]; 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; }