Submission #900355

#	Time	Username	Problem	Language	Result	Execution time	Memory
900355		Nelt	Tracks in the Snow (BOI13_tracks)	C++17	100 / 100	558 ms	236392 KiB

tracks

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

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

/* DEFINES */
#define S second
#define F first
#define ll long long
#define ull unsigned long long
#define ld long double
#define npos ULLONG_MAX
#define INF LLONG_MAX
#define vv(a) vector<a>
#define pp(a, b) pair<a, b>
#define pq(a) priority_queue<a>
#define qq(a) queue<a>
#define ss(a) set<a>
#define mm(a, b) map<a, b>
#define ump(a, b) unordered_map<a, b>
#define ANDROID                   \
    ios_base::sync_with_stdio(0); \
    cin.tie(0);                   \
    cout.tie(0);
#define elif else if
#define endl "\n"
#define allc(a) begin(a), end(a)
#define all(a) a, a + sizeof(a) / sizeof(a[0])
#define pb push_back
#define logi(a) __lg(a)
#define sqrt(a) sqrtl(a)
#define mpr make_pair
#define ins insert
using namespace std;
using namespace __gnu_pbds;
using namespace __cxx11;
typedef char chr;
typedef basic_string<chr> str;
template <typename T, typename key = less<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const ll N = 4005;
ll dist[N][N];
str s[N];
ll n, m;
ll dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};
bool can_go(ll x, ll y)
{
    return x > 0 and x <= n and y > 0 and y <= m and s[x][y] != '.';
}
void bfs()
{
    deque<pp(ll, ll)> q;
    for (ll i = 1; i <= n; i++)
        for (ll j = 1; j <= m; j++)
            dist[i][j] = 1e9;
    dist[1][1] = 0;
    q.pb(mpr(1, 1));
    while (!q.empty())
    {
        auto [x, y] = q.front();
        q.pop_front();
        for (ll k = 0; k < 4; k++)
            if (can_go(x + dx[k], y + dy[k]) and dist[x + dx[k]][y + dy[k]] > dist[x][y] + (s[x][y] != s[x + dx[k]][y + dy[k]]))
            {
                dist[x + dx[k]][y + dy[k]] = dist[x][y] + (s[x][y] != s[x + dx[k]][y + dy[k]]);
                if (s[x][y] != s[x + dx[k]][y + dy[k]])
                    q.pb(mpr(x + dx[k], y + dy[k]));
                else
                    q.push_front(mpr(x + dx[k], y + dy[k]));
            }
    }
}
void solve()
{
    cin >> n >> m;
    for (ll i = 1; i <= n; i++)
        cin >> s[i], s[i] = ' ' + s[i];
    bfs();
    ll ans = 0;
    for (ll i = 1; i <= n; i++)
    for (ll j = 1; j <= m; j++)
        if (s[i][j] != '.')
        ans = max(ans, dist[i][j]);
    cout << ans + 1 << endl;
}
/*

*/
int main()
{
    ANDROID
    // precomp();
    ll t = 1;
    // cin >> t;
    for (ll i = 1; i <= t; i++)
        // cout << "Case #" << i << ": ",
        solve();
    cerr << "\nTime elapsed : " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n";
}

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