Submission #489547

#	Time	Username	Problem	Language	Result	Execution time	Memory
489547		kiomi	Nautilus (BOI19_nautilus)	C++17	100 / 100	147 ms	720 KiB

nautilus

//#pragma GCC target ("avx2")
//#pragma GCC optimization ("Ofast")
//#pragma GCC optimization ("unroll-loops")

//#pragma comment(linker,"/stack:200000000")
//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

//order_of_key(k): Number of items strictly smaller than k .
//find_by_order(k): K-th element in a set (counting from zero).
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>

#define full(x,n) x,x+n+1
#define full(x) x.begin(),x.end()
#define finish return 0

#define putb push_back
#define f first
#define s second

//logx(a^n)=loga(a^n)/logx(a)
//logx(a*b)=logx(a)+logx(b)
//logx(y)=log(y)/log(x)
//logb(n)=loga(n)/loga(b)

#define ordered_set tree<ll,null_type,less<ll>,rb_tree_tag,tree_order_statistics_node_update>
#define putf push_front
#define gainb pop_back

//(a+b)^n=sum of C(n,i)*a^i*b^(n-i) 0<=i<=n
//(a-b)^n=sum of C(n,i)*a^i*b^(n-i) for even i-for odd i

#define gainf pop_front
#define len(x) (int)x.size()

// 1/b%mod=b^(m-2)%mod
// (a>>x)&1==0
// a^b=(a+b)-2(a&b)

typedef double db;
typedef long long ll;

//sum of squares n*(n+1)*(2n+1)/6
//sum of cubes [n*(n+1)/2]^2
//sum of squares for odds n*(4*n*n-1)/3
//sum of cubes for odds n*n*(2*n*n-1)

const int ary=500;
const ll mod=1e9+7;
const ll inf=1e18;

using namespace std;
using namespace __gnu_pbds;
bitset<ary> d[3][ary];
int r,c,m,ans;
char a;
string s;
int main(){
	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);
	cin>>r>>c>>m;
	for(int i=0;i<r;i++){
		for(int j=0;j<c;j++){
			cin>>a;
			d[1][i][j]=(a=='.');
			d[2][i][j]=d[1][i][j];
		}
	}
	cin>>s;
	for(int i=0;i<m;i++){
		int x=i%2;
		int y=1-x;
		for(int j=0;j<r;j++){
			if(s[i]=='?'||s[i]=='N'){
				if(j+1<r){
					d[x][j]|=d[y][j+1];
				}
			}
			if(s[i]=='?'||s[i]=='S'){
				if(j-1>=0){
					d[x][j]|=d[y][j-1];
				}
			}
			if(s[i]=='?'||s[i]=='W'){
				d[x][j]|=(d[y][j]>>1);
			}
			if(s[i]=='?'||s[i]=='E'){
				d[x][j]|=(d[y][j]<<1);
			}
		}
		for(int j=0;j<r;j++){
			d[x][j]&=d[2][j];
			if(i==m-1){
				ans+=d[x][j].count();
			}
			d[y][j].reset();
		}
	}
	cout<<ans;
}

Compilation message (stderr)

nautilus.cpp:15: warning: "full" redefined
   15 | #define full(x) x.begin(),x.end()
      | 
nautilus.cpp:14: note: this is the location of the previous definition
   14 | #define full(x,n) x,x+n+1
      |

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