Submission #1046620

#	Time	Username	Problem	Language	Result	Execution time	Memory
1046620		1L1YA	Maze (JOI23_ho_t3)	C++17	19 / 100	2095 ms	3076 KiB

Main

//1L1YA

#include<bits/stdc++.h>
using namespace std;

//#pragma GCC optimize ("O3,unrool-loops")
//#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")

typedef long long       ll;
typedef pair<ll,ll>     pll;
typedef pair<int,int>   pii;

#define Pb              push_back
#define F               first
#define S               second
#define endl            '\n'
#define sep             ' '
#define all(x)          x.begin(),x.end()
#define al(x,n)         x+1,x+n+1
#define SZ(x)           ((int)x.size())
#define lc              (id<<1)
#define rc              (lc|1)
#define mid             (l+r>>1)
#define dokme(x)        cout << x << endl, exit(0)
#define sik(x)          cout << x << endl;continue;
#define FastIO          ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define FileIO          freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll oo=1e18;
const int mod=1e9+7;
const int inf=1e9+111;
const int N=1e3+11;
const int lg=23;

int n,m,k,xs,ys,xt,yt,x[4]={1,-1,0,0},y[4]={0,0,1,-1},d[N][N];
string s[N];

void bfs(){
    deque<pii> dq;
    dq.Pb({xs,ys});
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            d[i][j]=inf;
    d[xs][ys]=0;
    while(!dq.empty()){
        int p=dq.front().F,q=dq.front().S;dq.pop_front();
        for(int i=0;i<4;i++)
            if(p+x[i] && p+x[i]<=n && q+y[i] && q+y[i]<=m && s[p+x[i]][q+y[i]]=='.' && d[p][q]<d[p+x[i]][q+y[i]]){
                d[p+x[i]][q+y[i]]=d[p][q];
                dq.push_front({p+x[i],q+y[i]});
            }
        for(int X=1;X<=n;X++)
            for(int Y=1;Y<=m;Y++)
                if(abs(X-p)<=k && abs(Y-q)<=k && (abs(X-p)<k || abs(Y-q)<k) && d[p][q]+1<d[X][Y]){
                    d[X][Y]=d[p][q]+1;
                    dq.Pb({X,Y});
                }
    }
}

int main(){
    FastIO

    cin >> n >> m >> k >> xs >> ys >> xt >> yt;
    for(int i=1;i<=n;i++){
        cin >> s[i];
        s[i]='#'+s[i];
    }
    bfs();
    cout << d[xt][yt] << endl;

    return 0;
}

• Subtask 1: 0 / 8
• Subtask 2: 19 / 19
• Subtask 3: 0 / 16
• Subtask 4: 0 / 19
• Subtask 5: 0 / 5
• Subtask 6: 0 / 19
• Subtask 7: 0 / 8
• Subtask 8: 0 / 6