Submission #1245524

#	Time	Username	Problem	Language	Result	Execution time	Memory
1245524		nanh0607	Tracks in the Snow (BOI13_tracks)	C++20	100 / 100	643 ms	110828 KiB

tracks

#include <bits/stdc++.h>
#include <cstdlib>
using namespace std;
using ll = long long;
using str = string;
using pii = pair<int, int>;
using pil = pair<int, ll>;
using pll = pair<ll, ll>;
#define FOR(i, l, r) for (int i=l; i<=r; i++)
#define FOR2(i, l, r) for (int i=l; i>=r; i--)
#define ALL(v) v.begin(), v.end()
#define fi first
#define se second
#define ce cout << endl
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define eb emplace_back
#define ef emplace_front
#define ep emplace
const ll MOD = 998244353;
const int INF = 1e9;
const int LOG = 60;
const double MAX = 1e9;
int dx[4] = {1, -1, 0, 0};
int dy[4] = {0, 0, 1, -1};
// north, south, east, weast

void solve() {
    int n, m; cin >> n >> m;
    vector<vector<char>> meadow (n, vector<char> (m));
    FOR(i, 0, n-1) FOR(j, 0, m-1) cin >> meadow[i][j];

    function<bool(int x, int y)> inside = [&] (int x, int y) {
        return x >= 0 && x < n && y >= 0 && y < m && meadow[x][y] != '.';
    };

    vector<vector<int>> depth(n, vector<int> (m, -1));
    deque<pii> dq;

    depth[0][0] = 0;
    dq.eb(0, 0);
    int res = 0;

    while (!dq.empty()) {
        auto [x1, y1] = dq.back(); dq.pob();
        res = max(res, depth[x1][y1]);

        FOR(i, 0, 3) {
            int x2 = x1 + dx[i], y2 = y1 + dy[i];
            if (!inside(x2, y2) || depth[x2][y2] != -1) continue;
            if (meadow[x2][y2] == meadow[x1][y1]) {
                depth[x2][y2] = depth[x1][y1];
                dq.eb(x2, y2);
            } else {
                depth[x2][y2] = depth[x1][y1]+1;
                dq.ef(x2, y2);
            }
        }
    }
    cout << res+1;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
    // freopen("movie.in", "r", stdin);
    // freopen("movie.out", "w", stdout);
    int t = 1;
    // cin >> t;
    while (t--) solve();
}

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