이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define ll long long
#define endl "\n"
using namespace std;
using namespace __gnu_pbds;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T, typename key = less_equal<T>>
using ordered_set = tree<T, null_type, key, rb_tree_tag, tree_order_statistics_node_update>;
const ll N = 4005;
string s[N];
ll dist[N][N], dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1}, n, m;
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()
{
for (ll i = 1; i <= n; i++)
for (ll j = 1; j <= m; j++)
dist[i][j] = 1e9;
dist[1][1] = 1;
deque<pair<ll, ll>> q;
q.push_back(make_pair(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.push_back(make_pair(x + dx[k], y + dy[k]));
else q.push_front(make_pair(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 << endl;
}
int main()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
ll t = 1;
// precomp();
// cin >> t;
for (ll cs = 1; cs <= t; cs++)
solve();
// cerr << "\nTime elapsed: " << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n";
}
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