結果
問題 | No.20 砂漠のオアシス |
ユーザー |
|
提出日時 | 2023-09-27 19:36:10 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
WA
|
実行時間 | - |
コード長 | 3,545 bytes |
コンパイル時間 | 4,492 ms |
コンパイル使用メモリ | 266,548 KB |
最終ジャッジ日時 | 2025-02-17 02:40:01 |
ジャッジサーバーID (参考情報) |
judge1 / judge2 |
(要ログイン)
ファイルパターン | 結果 |
---|---|
other | AC * 16 WA * 5 |
ソースコード
#include <bits/stdc++.h>using namespace std;#include <atcoder/all>#include <time.h>using namespace atcoder;using ll = long long;using vll = vector<ll>;using vvll = vector<vll>;using vvvll = vector<vvll>;using vb = vector<bool>;using vvb = vector<vb>;using vvvb = vector<vvb>;#define all(A) A.begin(),A.end()#define rep(i, n) for (ll i = 0; i < (ll) (n); i++)template<class T>bool chmax(T& p, T q, bool C = 1) {if (C == 0 && p == q) {return 1;}if (p < q) {p = q;return 1;}else {return 0;}}template<class T>bool chmin(T& p, T q, bool C = 1) {if (C == 0 && p == q) {return 1;}if (p > q) {p = q;return 1;}else {return 0;}}ll modPow(long long a, long long n, long long p) {if (n == 0) return 1; // 0乗にも対応する場合if (n == 1) return a % p;if (n % 2 == 1) return (a * modPow(a, n - 1, p)) % p;long long t = modPow(a, n / 2, p);return (t * t) % p;}ll gcd(ll(a), ll(b)) {if (a == 0)return b;if (b == 0)return a;ll c = a;while (a % b != 0) {c = a % b;a = b;b = c;}return b;}ll sqrtz(ll N) {ll L = 0;ll R = sqrt(N) + 10000;while (abs(R - L) > 1) {ll mid = (R + L) / 2;if (mid * mid <= N)L = mid;else R = mid;}return L;}using mint = modint998244353;using vm = vector<mint>;using vvm = vector<vm>;using vvvm = vector<vvm>;vector<mint> fact, factinv, inv;const ll mod = 998244353;void prenCkModp(ll n) {fact.resize(n + 5);factinv.resize(n + 5);inv.resize(n + 5);fact[0] = fact[1] = 1;factinv[0] = factinv[1] = 1;inv[1] = 1;for (ll i = 2; i < n + 5; i++) {fact[i] = (fact[i - 1] * i);inv[i] = (mod - ((inv[mod % i] * (mod / i))));factinv[i] = (factinv[i - 1] * inv[i]);}}mint nCk(ll n, ll k) {if (n < k || k < 0) return 0;return (fact[n] * ((factinv[k] * factinv[n - k])));}bool DEB = 1;template<class T>vector<vector<T>> mul(vector<vector<T>> A,vector<vector<T>> B){ll N=A.size();vector<vector<T>> res(N,vector<T> (N,0));rep(i,N)rep(j,N)rep(k,N)res[i][j]+=A[i][k]*B[k][j];return res;}int main() {ll N,V,OX,OY;cin>>N>>V>>OX>>OY;OX--;OY--;vvvll D(2,vvll(N,vll(N,1e18)));vvll L(N,vll(N));rep(i,N)rep(j,N)cin>>L[i][j];vll dx={1,-1,0,0};vll dy={0,0,1,-1};rep(p,2){priority_queue<pair<ll,pair<ll,ll>>,vector<pair<ll,pair<ll,ll>>>,greater<pair<ll,pair<ll,ll>>>> Q;vector<vector<bool>> seen(N,vb(N,false));if(p==0){Q.push({0,{0,0}});D[p][0][0]=0;}else {Q.push({0,{N-1,N-1}});D[p][N-1][N-1]=0;}while(!Q.empty()){auto cyx=Q.top();Q.pop();ll c=cyx.first;ll y=cyx.second.first;ll x=cyx.second.second;rep(d,4){ll ny=y+dy[d];ll nx=x+dx[d];if(min(ny,nx)<0||max(ny,nx)>=N)continue;if(seen[ny][nx])continue;if(chmin(D[p][ny][nx],c+L[ny][nx])){Q.push({D[p][ny][nx],{ny,nx}});}}}}bool OK=0;if(D[0][N-1][N-1]<V)OK=1;if(OX!=-1){ll Z=V-D[0][OX][OY];Z*=2;if(D[1][OX][OY]<Z)OK=1;}cout<<(OK?"YES":"NO")<<endl;}