Submission #914956

#	Time	Username	Problem	Language	Result	Execution time	Memory
914956		HorizonWest	Dango Maker (JOI18_dango_maker)	C++17	100 / 100	1489 ms	18800 KiB

dango_maker

#include <bits/stdc++.h>
using namespace std;

#define endl '\n'
#define db double
#define ll __int128
#define int long long
#define pb push_back
#define fs first
#define sd second
#define Mod long(1e9 + 7)
#define all(x) x.begin(), x.end()
#define unvisited long(-1)
#define Eps double(1e-9)
#define _for(i, n) for(int i = 0; i < (n); i++)
#define dbg(x) cout << #x ": " << x << endl;

const int Max = (1LL << 32), Inf = 1e9 + 7;

void print(bool x) { cout << (x ? "YES" : "NO") << endl; }

string tostring (ll x)
{
    string ans = "";
    while(x > 0)
    {
        ans += (x % 10 + '0');
        x /= 10;
    }
    reverse(all(ans));
    return ans;
}

void solve()
{
    int n, m, ans = 0; cin >> n >> m; 
    vector <vector<char>> v(n, vector <char> (m, 0));
    for(auto& u : v) for(auto& t : u) cin >> t;

    for(int i = 1; i < n+m; i++)
    {
        vector <vector<int>> dp(n+7, vector <int> (3, 0));
        vector <vector<char>> s(n+7, vector <char> (3, '.'));

        for(int j = 0; j < n; j++)
        {
            for(int k = 0; k < 3; k++)
            {
                int x = j, y = i - j + k - 1;
                if(x < 0 || x > n - 1 || y < 0 || y > m - 1) continue;
                s[j][k] = v[x][y];
            }
        }

        for(int j = 0; j <= n+3; j++)
        {
            for(int k = 0; k < 3; k++)
            {
                dp[j+1][0] = max(dp[j+1][0], dp[j][k]);
                if(s[j][0] == 'R' && s[j][1] == 'G' && s[j][2] == 'W')
                    dp[j+1][1] = max(dp[j][0], dp[j][1]) + 1;
                if(j != 0)
                {
                    if(s[j-1][0] == 'R' && s[j][1] == 'G' && s[j+1][2] == 'W')
                        dp[j+1][2] = max(dp[j][2], dp[j][0]) + 1; 
                }
            }
        }
      
        ans += dp[n+3][0];
    }

    cout << ans << endl;
}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int Q = 1; //cin >> Q;

    while (Q--)
    {
        solve();
    }

    return 0;
}

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