제출 #1245405

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1245405		mohamedyehia	Tracks in the Snow (BOI13_tracks)	C++20	0 / 100	849 ms	229160 KiB

tracks

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

typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;

#define ffr(i, s, n) for (int i = s; i < n; i++)

vector<vll> depth;
ll ans = 1;
vll dx = {-1, 1, 0, 0};
vll dy = {0, 0, -1, 1};

bool check_in_side(ll i, ll j, ll n, ll m)
{
    return (i >= 0 && i < n && j >= 0 && j < m);
}

void bfs(ll n, ll m, vector<string> &snow)
{
    // build the queue
    deque<pll> q;
    // strarting from node 0 , 0;
    q.push_front(make_pair(0, 0));
    // its depth is 1; but i have a question what about if there is no animal ?
    // how we deal with '.'
    depth[0][0] = 1;
    while (!q.empty())
    {
        // pop the forntier;
        pll forntier = q.front();
        q.pop_front();

        // make the code easier by taking the x, and y
        ll x = forntier.first;
        ll y = forntier.second;
        // update ans if needed
        ans = max(ans, depth[x][y]);
        // go through each child
        for (int i = 0; i < 4; i++)
        {
            ll nx = x + dx[i];
            ll ny = y + dy[i];
            // if they are not visited and inside the boundaries queue them
            if (check_in_side(nx, ny, n, m) && depth[nx][ny] == 0)
            {
                // give high proirity for the similar nodes
                if (snow[nx][ny] == snow[x][y])
                {
                    q.push_front(make_pair(nx, ny));
                    depth[nx][ny] = depth[x][y];
                }
                else
                // lowest priority
                {
                    q.push_back(make_pair(nx, ny));
                    depth[nx][ny] = depth[x][y] + 1;
                }
            }
        }
    }
}
void solve()
{
    ll n, m;
    cin >> n >> m;

    depth.assign(n, vector<ll>(m));
    vector<string> snow(n);
    ffr(i, 0, n)
    {
        cin >> snow[i];
    }
    
    bfs(n, m , snow);
    cout << ans - 1 << "\n";
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    // freopen("milkvisits.out", "r", stdin);
    solve();
    return 0;
}

• Subtask 1: 0 / 30
• Subtask 2: 0 / 70