Submission #210323

#	Time	Username	Problem	Language	Result	Execution time	Memory
210323		Blerargh	Nautilus (BOI19_nautilus)	C++17	100 / 100	236 ms	760 KiB

nautilus

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

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> ii;
typedef pair<ld,ld> id;

#define FOR(i, a, b) for(int i=(a); i<=(b); i++)
#define ROF(i, a, b) for(int i=(a); i>=(b); i--)
#define MEM(x, v) memset(x, v, sizeof(x))
#define FILL(x, n, v) fill(x, x+n, v);
#define ALL(x) x.begin(), x.end()
#define SORT(x) sort((x).begin(), (x).end())
#define CMPSORT(x, cp) sort((x).begin(), (x).end(), cp)
#define FAST ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);

#define f first
#define s second
#define ins insert
#define e emplace
#define eb emplace_back
#define ef emplace_front
#define p push
#define pf push_front
#define pb push_back
#define mp make_pair
#define ub upper_bound
#define lb lower_bound
#define ft front
#define bk back
#define pp pop
#define ppb pop_back
#define ppf pop_front

#define db cout<<"YEET\n";
#define ct(x) cout<<x<<'\n';

const ll MOD = 1e9+7; //998244353
const ll MAXN = 2e5+5;
const ll INF = 1e18;
const ld PI = acos((ld)-1);

int main(){
    FAST
    ll r, c, k;
    bitset<505> dp[505], sea[505], prev[505];
    cin >> r >> c >> k;
    FOR(i,1,r){
        FOR(j,1,c){
            char ch;
            cin >> ch;
            if (ch == '.'){
                sea[i][j]=1;
                dp[i][j]=1;
                prev[i][j]=1;
            } else {
                sea[i][j]=0;
                dp[i][j]=0;
                prev[i][j]=0;
            }
        }
    }

    FOR(i,1,k){
        char ch;
        cin >> ch;

        FOR(j,1,r){
            if (ch == 'N'){
                dp[j] = prev[j+1]&sea[j];
            } else if (ch == 'E'){
                dp[j] = (prev[j]<<1)&sea[j];
            } else if (ch == 'S'){
                dp[j] = prev[j-1]&sea[j];
            } else if (ch == 'W'){
                dp[j] = (prev[j]>>1)&sea[j];
            } else {
                dp[j] = (prev[j-1] | prev[j+1] | prev[j]>>1 | prev[j]<<1)&sea[j];
            }
        }
        FOR(j,1,r) prev[j]=dp[j];
    }

    ll res=0;
    FOR(i,1,r) res+=dp[i].count();
    cout << res;
}

• Subtask 1: 29 / 29
• Subtask 2: 37 / 37
• Subtask 3: 34 / 34