We observe that in order to reach a cell in the grid, we need to choose a path with the minimum number of character changes. Therefore, we use Dijkstra’s algorithm; however, since the edge weights are only 0 or 1, it can be simplified into BFS (more precisely, 0-1 BFS)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define ld long double
#define FOR(i,l,r) for (int i = (l), _r = (r); i <= _r; i++)
#define REP(i,l,r) for (int i = (l), _r = (r); i < _r; i++)
#define FORN(i,r,l) for (int i = (r), _l = (l); i >= _l; i--)
#define MASK(x) (1LL << (x))
#define BIT(x,i) (((x) >> (i)) & 1)
#define sz(x) (int)x.size()
#define all(v) begin((v)), end((v))
#define allVector(v, n) begin((v)) + 1, begin((v)) + (n) + 1
#define segleft (id << 1)
#define segright (id << 1 | 1)
#define tcT template <class T
tcT> bool minimize(T& a, const T& b) { return a > b ? a = b, 1 : 0; }
tcT> bool maximize(T& a, const T& b) { return a < b ? a = b, 1 : 0; }
#define endl '\n'
#define fi first
#define se second
#define pf push_front
#define pb push_back
#define lb lower_bound
#define ub upper_bound
const int MOD = 1e9 + 7;

const int N = 4e3 + 1;
const int dx[] = {1, -1, 0, 0};
const int dy[] = {0, 0, 1, -1}; 
char board[N][N];
int dist[N][N];
int n, m;

bool check(int x, int y) {
    return x >= 1 && x <= n && y >= 1 && y <= m && board[x][y] != '.' && dist[x][y] == 0;
}

void solve() {
    cin >> n >> m;

    FOR(i, 1, n) {
        FOR(j, 1, m) {
            cin >> board[i][j];
        }
    }

    deque<pair<int,int>> q;
    dist[1][1] = 1;
    q.pb({1, 1});
    int ans = 1;

    while (!q.empty()) {
        auto[x, y] = q.front(); q.pop_front();

        maximize(ans, dist[x][y]);

        REP(i, 0, 4) {
            int nx = x + dx[i], ny = y + dy[i];

            if (!check(nx, ny)) continue;

            if (board[nx][ny] == board[x][y]) {
                dist[nx][ny] = dist[x][y];
                q.pf({nx, ny});
            } else {
                dist[nx][ny] = dist[x][y] + 1;
                q.pb({nx, ny});
            }
        }
    }

    cout << ans << endl;
}