이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cmath>
#include <cstring>
#include <deque>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef vector<pair<int, int>> vpi;
#define x first
#define fastio() \
ios::sync_with_stdio(false); \
cin.tie(0); \
cout.tie(0)
#define y second
#define PB push_back
#define MP make_pair
#define REP(i, a, b) for (int i = a; i < b; i++)
int mod = 1e9 + 7;
int mod_inverse_2 = 500000004; // Modular inverse of 2 modulo (10^9 + 7), also
// the number has to be in long long otherwise
// if the number is int need to mutiply by 1LL
const int N = 100009;
int dx[] = {0, 0, -1, 1};
int dy[] = {1, -1, 0, 0};
ll bound = 1e18;
ll H, W;
vector<vector<char>> grid;
vector<vector<ll>> depths;
void solve()
{
cin >> H >> W;
grid.resize(H + 1);
depths.resize(H + 1);
REP(i, 0, H + 1)
{
grid[i].resize(W + 1);
depths[i].resize(W + 1, 0);
}
REP(i, 1, H + 1)
{
REP(j, 1, W + 1)
{
cin >> grid[i][j];
}
}
deque<pair<ll, ll>> q;
depths[1][1] = 1;
q.push_back({1, 1});
ll ans = 0;
while (!q.empty())
{
pair<ll, ll> curr = q.front();
q.pop_front();
ans = max(ans, depths[curr.x][curr.y]);
REP(k, 0, 4)
{
if (curr.x + dy[k] < 1 || curr.y + dx[k] < 1 ||
curr.x + dy[k] > H || curr.y + dx[k] > W)
{
continue;
}
pair<ll, ll> nxt = {curr.x + dy[k], curr.y + dx[k]};
if (grid[nxt.x][nxt.y] == '.')
{
continue;
}
if (depths[nxt.x][nxt.y] != 0)
{
continue;
}
if (grid[nxt.x][nxt.y] == grid[curr.x][curr.y])
{
depths[nxt.x][nxt.y] = depths[curr.x][curr.y];
q.push_front(nxt);
}
else
{
depths[nxt.x][nxt.y] = depths[curr.x][curr.y] + 1;
q.push_back(nxt);
}
}
}
cout << ans << "\n";
}
int main()
{
fastio();
solve();
return 0;
}
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