Submission #555729

#	Time	Username	Problem	Language	Result	Execution time	Memory
555729		SavicS	Nautilus (BOI19_nautilus)	C++17	100 / 100	178 ms	1000 KiB

nautilus

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <bits/stdc++.h>
 
using namespace std;
using namespace __gnu_pbds;
 
typedef long long ll;
typedef long double ld;
 
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef pair<ld,ld> pdd;
 
#define ff(i,a,b) for(int i = a; i <= b; i++)
#define fb(i,b,a) for(int i = b; i >= a; i--)
#define trav(a,x) for (auto& a : x)
 
#define sz(a) (int)(a).size()
#define pb push_back
#define fi first
#define se second
#define lb lower_bound
#define ub upper_bound
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
 
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

template<typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

// os.order_of_key(k) the number of elements in the os less than k
// *os.find_by_order(k)  print the k-th smallest number in os(0-based)

const int mod = 1000000007;
const int inf = 1e9 + 5;
const int mxN = 505; 

const int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

int n, m, k;
string A[mxN];
string s;

bool dp[mxN][mxN][2];


int main() {
    cin.tie(0)->sync_with_stdio(0);

    cin >> n >> m >> k;
    ff(i,0,n - 1)cin >> A[i];
    cin >> s;

    // I - koja su polja voda
    // P - prethodni korak

	bitset<mxN> I[n], P[n];
    ff(i,0,n - 1){
    	ff(j,0,m - 1){
    		if(A[i][j] == '.'){
    			I[i][j] = P[i][j] = 1;
    		}
    	}
    }

    ff(l,0,k - 1){

    	bitset<mxN> nwP[n];

    	if(s[l] == 'N' || s[l] == '?'){
    		ff(i,0,n - 2){
    			nwP[i] |= P[i + 1];
    		}
    	}

    	if(s[l] == 'S' || s[l] == '?'){
    		ff(i,1,n - 1){
    			nwP[i] |= P[i - 1];
    		}
    	}

    	if(s[l] == 'W' || s[l] == '?'){
    		ff(i,0,n - 1){
    			nwP[i] |= (P[i] >> 1);
            }
    	}

    	if(s[l] == 'E' || s[l] == '?'){
            ff(i,0,n - 1){
                nwP[i] |= (P[i] << 1);

            }
    	}

        ff(i,0,n - 1)P[i] = (nwP[i] & I[i]);

    }


    int rez = 0;
    ff(i,0,n - 1)rez += P[i].count();

    cout << rez << '\n';

    return 0;
}
/*
 
5 9 7
...##....
..#.##..#
..#....##
.##...#..
....#....
WS?EE?? 
 
// probati bojenje sahovski
*/
 
 
 
 
 

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