結果
問題 | No.20 砂漠のオアシス |
ユーザー |
|
提出日時 | 2024-09-27 15:26:55 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,041 bytes |
コンパイル時間 | 5,842 ms |
コンパイル使用メモリ | 291,484 KB |
最終ジャッジ日時 | 2025-02-24 13:24:40 |
ジャッジサーバーID (参考情報) |
judge2 / judge2 |
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ファイルパターン | 結果 |
---|---|
other | AC * 16 WA * 5 |
ソースコード
// #pragma GCC target("avx2")#pragma GCC optimize("O3")#pragma GCC optimize("unroll-loops")#include <bits/stdc++.h>#include <atcoder/all>using namespace std;using namespace atcoder;using mint = modint998244353;// using mint = modint1000000007;using ll = long long;using ull = unsigned long long;using ld = long double;using pii = pair<int, int>;using pll = pair<ll, ll>;using T = tuple<int, int, int>;using G = vector<vector<int>>;#define rep(i, n) for (ll i = 0; i < (n); ++i)#define rep2(i, a, b) for (ll i = a; i < (b); ++i)#define rrep2(i, a, b) for (ll i = a-1; i >= (b); --i)#define rep3(i, a, b, c) for (ll i = a; i < (b); i+=c)#define rng(a) a.begin(),a.end()#define rrng(a) a.rbegin(),a.rend()#define popcount __builtin_popcount#define popcountll __builtin_popcountll#define fi first#define se second#define UNIQUE(v) sort(rng(v)), v.erase(unique(rng(v)), v.end())#define MIN(v) *min_element(rng(v))#define MAX(v) *max_element(rng(v))template<class T> bool chmin(T &a,T b){if(a>b){a=b;return 1;}else return 0;}template<class T> bool chmax(T &a,T b){if(a<b){a=b;return 1;}else return 0;}template<class T> void printv(vector<T> &v){rep(i,v.size())cout<<v[i]<<" \n"[i==v.size()-1];}template<class T> void printvv(vector<vector<T>> &v){rep(i,v.size())rep(j,v[i].size())cout<<v[i][j]<<" \n"[j==v[i].size()-1];cout<<endl;}const ll dx[] = {0, 1, 0, -1};const ll dy[] = {1, 0, -1, 0};const ll dxx[] = {0, 1, 0, -1, 1, -1, 1, -1};const ll dyy[] = {1, 0, -1, 0, 1, 1, -1, -1};const ll LINF = 1001002003004005006ll;const int INF = 1001001001;int main(){ios::sync_with_stdio(false);cin.tie(nullptr);int n, h; cin >> n >> h;int ox, oy; cin >> ox >> oy; ox--, oy--;vector<vector<pii>> g(n*n);rep(i, n)rep(j, n){int w; cin >> w;rep(t, 4){int ni = i+dx[t], nj = j+dy[t];if (ni < 0 || ni >= n || nj < 0 || nj >= n) continue;g[ni*n+nj].emplace_back(i*n+j, w);}}auto dijkstra = [&](int sx, int sy, int gx, int gy, int &h, int f)-> bool{vector<int> dist(n*n, INF); dist[sx*n+sy] = 0;priority_queue<pii, vector<pii>, greater<pii>> pq;pq.emplace(0, sx*n+sy);while(!pq.empty()){auto [c, u] = pq.top(); pq.pop();if (dist[u] < c) continue;for(auto [v, w]: g[u]){if (dist[v] > dist[u]+w && dist[u]+w < h){dist[v] = dist[u]+w;pq.emplace(dist[v], v);}}}if (f) h -= dist[gx*n+gy];return dist[gx*n+gy] != INF;};if (dijkstra(0, 0, n-1, n-1, h, 0)){cout << "YES" << endl;return 0;}if (ox != -1){if (!dijkstra(0, 0, ox, oy, h, 1)){cout << "NO" << endl;return 0;}h *= 2;if (dijkstra(ox, oy, n-1, n-1, h, 0)){cout << "YES" << endl;return 0;}}cout << "NO" << endl;return 0;}