Submission #333582

#	Time	Username	Problem	Language	Result	Execution time	Memory
333582		mohamedsobhi777	Dango Maker (JOI18_dango_maker)	C++14	13 / 100	1 ms	512 KiB

dango_maker

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

#pragma GCC optimize("-Ofast")
//#pragma GCC optimize("trapv")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,sse4.2,popcnt,abm,mmx,avx2,tune=native")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-funroll-loops")

using namespace std;
using namespace __gnu_pbds;

#define vi vector<int>
#define vii vector<pair<int, int>>
#define pii pair<int, int>
#define pll pair<ll, ll>
#define loop(_) for (int __ = 0; __ < (_); ++__)
#define forn(i, n) for (int i = 0; i < n; ++i)
#define pb push_back
#define f first
#define sz(_) ((int)_.size())
#define all(_) _.begin(), _.end()
#define uni(_) unique(_)
#define lb lower_bound
#define ub upper_bound
#define si set<int>
#define ms multiset<int>
#define qi queue<int>
#define pq prioriry_queue<int>

using ll = long long;
using ld = long double;

const int N = 3000 + 7;
const ll mod = 1e9 + 7;
const ll inf = 2e18;

auto ra = [] {char *p = new char ; delete p ; return ll(p) ; };
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count() * (ra() | 1));
typedef tree<pii, null_type, less<pii>, rb_tree_tag, tree_order_statistics_node_update> os;

int n, m;
string s[N];
bool on[N][N];

int main()
{
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        //freopen("in.in", "r", stdin);
        cin >> n >> m;
        for (int i = 0; i < n; ++i)
                cin >> s[i];
        int ans = 0;
        for (int i = 0; i < n; ++i)
        {
                for (int j = 0; j < m; ++j)
                {
                        if (i > 1 && s[i][j] == 'W' && s[i - 1][j] == 'G' && s[i - 2][j] == 'R' && (on[i][j] + on[i - 1][j] + on[i - 2][j] == 0))
                        {
                                ++ans;
                                on[i][j] = on[i - 1][j] = on[i - 2][j] = 1;
                        }
                        if (j > 1 && s[i][j] == 'W' && s[i][j - 1] == 'G' && s[i][j - 2] == 'R' && (on[i][j] + on[i][j - 1] + on[i][j - 2] == 0))
                        {
                                ++ans;
                                on[i][j] = on[i][j - 1] = on[i][j - 2] = 1;
                        }
                }
        }
        
        cout << ans;
        return 0;
}

• Subtask 1: 13 / 13
• Subtask 2: 0 / 20
• Subtask 3: 0 / 67