結果

問題 No.20 砂漠のオアシス
ユーザー shihu
提出日時 2024-09-27 15:35:30
言語 C++17
(gcc 13.3.0 + boost 1.87.0)
結果
AC  
実行時間 27 ms / 5,000 ms
コード長 3,038 bytes
コンパイル時間 5,682 ms
コンパイル使用メモリ 292,140 KB
最終ジャッジ日時 2025-02-24 13:25:01
ジャッジサーバーID
（参考情報） 		judge5 / judge1
このコードへのチャレンジ
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ファイルパターン 結果
other AC * 21
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ソースコード

diff #
プレゼンテーションモードにする

// #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 >> oy >> ox; 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, 1)){
            cout << "YES" << endl;
            return 0;
        }
    }
    cout << "NO" << endl;
    return 0;
}
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