#include #include #include #include #include #include using namespace std; typedef pair PI; typedef pair V; const int d[5] = { 0, 1, 0, -1, 0 }; int cost[201][201]; int dijkstra(PI s, PI t, int n){ int dist[202][202]; for (int i = 1; i <= n; i++)for (int j = 1; j <= n; j++)dist[i][j] = -1; for (int i = 0; i <= n + 1; i++)dist[0][i] = dist[i][0] = dist[n + 1][i] = dist[i][n + 1] = 100000; priority_queue, greater > que; que.push(make_pair(0, s)); while (que.empty() == false){ PI v = que.top().second; int di = que.top().first; que.pop(); if (dist[v.first][v.second] >= 0)continue; dist[v.first][v.second] = di; for (int i = 0; i<4; i++){ PI p = v; p.first += d[i]; p.second += d[i + 1]; if (dist[p.first][p.second] >= 0)continue; que.push(make_pair(di + cost[p.first][p.second], p)); } } return dist[t.first][t.second]; } int main(){ int N; int V; PI O; cin >> N >> V >> O.second >> O.first; for (int i = 1; i <= N; i++) for (int j = 1; j <= N; j++) cin >> cost[i][j]; auto st = make_pair(1, 1); auto ed = make_pair(N, N); if (dijkstra(st, ed, N)0 && U - dijkstra(O, ed, N)>0) cout << "YES" << endl; else cout << "NO" << endl; } return 0; }