Submission #498000

#	Time	Username	Problem	Language	Result	Execution time	Memory
498000		AQ0212	Nautilus (BOI19_nautilus)	C++17	66 / 100	88 ms	17932 KiB

nautilus

#include <iostream>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <sstream>
#include <cstring>
#include <bitset>

// #pragma GCC optimize("inline")
// #pragma GCC optimize("-fgcse,-fgcse-lm")
// #pragma GCC optimize("-ftree-pre,-ftree-vrp")
// #pragma GCC optimize("-ffast-math")
// #pragma GCC optimize("-fipa-sra")
// #pragma GCC optimize("-fpeephole2")
// #pragma GCC optimize("-fsched-spec")
// #pragma GCC optimize("Ofast,no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
// #pragma GCC optimize("unroll-loops")

#define ll long long int
#define pb push_back
#define pll pair < ll , ll >
#define fi first
#define se second
#define all(x) x.begin(), x.end()

using namespace std;
 
ll inf2 = 3e18;
 
ll dp[ 103 ][ 103 ][ 103 ], ans;
// bitset < 100111 > bst[ n ];
string s;
 
int main (){
    ll n, m, t;
	char c;
 
	cin >> n >> m >> t;
 
	for(int i = 1; i <= n; i ++){
		for(int j = 1; j <= m; j ++){
			cin >> c;
			if(c == '.')
				dp[ 0 ][ i ][ j ] = 1;
		}
	}
 
	cin >> s;
	// s = '#' + s;
	// reverse(all(s));

	for(int k = 1; k <= t; k ++){
		for(int i = 1; i <= n; i ++){
			for(int j = 1; j <= m; j ++){
				if(!dp[ 0 ][ i ][ j ]){
					dp[ k ][ i ][ j ] = 0;
					continue;
				}
				if(s[ k - 1 ] == 'N' && i + 1 <= n)
					dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i + 1 ][ j ];
				else if(s[ k - 1 ] == 'S' && i - 1 > 0)
					dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i - 1 ][ j ];
				else if(s[ k - 1 ] == 'W' && j + 1 <= m)
					dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i ][ j + 1 ];
				else if(s[ k - 1 ] == 'E' && j - 1 > 0)
					dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i ][ j - 1 ];
				else if(s[ k - 1 ] == '?'){
					if(i + 1 <= n)
						dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i + 1 ][ j ];
					if(i - 1 > 0)
						dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i - 1 ][ j ];
					if(j + 1 <= m)
						dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i ][ j + 1 ];
					if(j - 1 > 0)
						dp[ k ][ i ][ j ] |= dp[ k - 1 ][ i ][ j - 1 ];
				}
			}
		}
	}

	// for(int k = 0; k <= t; k ++){
		for(int i = 1; i <= n; i ++){
			for(int j = 1; j <= m; j ++){
				ans += dp[ t ][ i ][ j ];
				// cout << dp[ k ][ i ][ j ] << " ";
			}
			// cout << endl;
		}
		// cout << endl;
	// }

	cout << ans;
}
 
/*
 
5 9 7
...##....
..#.##..#
..#....##
.##...#..
....#....
WS?EE??
 
3 3 3
#..
.#.
..#

*/

• Subtask 1: 29 / 29
• Subtask 2: 37 / 37
• Subtask 3: 0 / 34