Submission #752216

#	Time	Username	Problem	Language	Result	Execution time	Memory
752216		GrindMachine	Nautilus (BOI19_nautilus)	C++17	100 / 100	45 ms	672 KiB

nautilus

// Om Namah Shivaya

#pragma GCC optimize("O3,unroll-loops")
#pragma GCC target("avx2")

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

using namespace std;
using namespace __gnu_pbds;

template<typename T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
typedef long long int ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define pb push_back
#define endl '\n'
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#define ff first
#define ss second
#define conts continue
#define ceil2(x, y) ((x + y - 1) / (y))
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "Yes" << endl
#define no cout << "No" << endl

#define rep(i, n) for(int i = 0; i < n; ++i)
#define rep1(i, n) for(int i = 1; i <= n; ++i)
#define rev(i, s, e) for(int i = s; i >= e; --i)
#define trav(i, a) for(auto &i : a)

template<typename T>
void amin(T &a, T b) {
    a = min(a, b);
}

template<typename T>
void amax(T &a, T b) {
    a = max(a, b);
}

#ifdef LOCAL
#include "debug.h"
#else
#define debug(x) 42
#endif

/*

refs:
edi

almost solved it on my own (got the dp idea and also optimized with bitset), but made a mistake:
i was initially iterating over all m and setting dp[i][j] to 0 if a[i][j] == '#', which made the complexity O(nmk) 
replaced this part also with bitset after reading edi and it passed

*/

const int MOD = 1e9 + 7;
const int N = 500 + 5;
const int inf1 = int(1e9) + 5;
const ll inf2 = ll(1e18) + 5;

void solve(int test_case)
{
    ll n,m,k; cin >> n >> m >> k;
    char a[n+5][m+5];
    rep(i,n+5) rep(j,m+5) a[i][j] = '#';
    rep1(i,n) rep1(j,m) cin >> a[i][j];
    string s; cin >> s;
    s = "$" + s;

    bitset<N> dp1[n+5], dp2[n+5], good[n+5];

    rep1(i,n){
        rep1(j,m){
            if(a[i][j] == '.'){
                dp1[i][j] = 1;
                good[i][j] = 1;
            }
        }
    }

    rep1(p,k){
        char ch = s[p];

        rep1(i,n){
            if(ch == '?'){
                dp2[i] = (dp1[i-1] | dp1[i+1] | (dp1[i] << 1) | (dp1[i] >> 1));
            }
            else{
                if(ch == 'N') dp2[i] = dp1[i+1];
                if(ch == 'S') dp2[i] = dp1[i-1];
                if(ch == 'E') dp2[i] = (dp1[i] << 1);
                if(ch == 'W') dp2[i] = (dp1[i] >> 1);
            }

            dp2[i] &= good[i];
        }

        rep1(i,n){
            dp1[i] = dp2[i];
        }
    }

    ll ans = 0;
    rep1(i,n){
        rep1(j,m){
            ans += dp1[i][j];
        }
    }

    cout << ans << endl;
}

int main()
{
    fastio;

    int t = 1;
    // cin >> t;

    rep1(i, t) {
        solve(i);
    }

    return 0;
}

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