【題解】Codeforces 1350E. Orac and Game of Life

#include <iostream>
#include <queue>
using namespace std;
 
int n, m, t, d[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
int g[1005][1005], r[1005][1005], a, b;
long long c;
string s;
queue <pair<int, int>> q;
 
int main() {
    cin >> n >> m >> t;
    for (int i = 0; i < n; i++){
        cin >> s;
        for (int j = 0; j < m; j++){
            g[i][j] = s[j]-'0';
            r[i][j] = -1;
        }
    }
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            for (int k = 0; k < 4; k++){
                if (0 <= i+d[k][0] && i+d[k][0] < n && 0 <= j+d[k][1] && j+d[k][1] < m && g[i+d[k][0]][j+d[k][1]] == g[i][j]){
                    q.push({i, j});
                    r[i][j] = 0;
                    break;
                }
            }
        }
    }
    while (!q.empty()){
        pair<int, int> p = q.front();
        q.pop();
        for (int k = 0; k < 4; k++){
            if (0 <= p.first+d[k][0] && p.first+d[k][0] < n && 0 <= p.second+d[k][1] && p.second+d[k][1] < m){
                if (r[p.first+d[k][0]][p.second+d[k][1]] == -1){
                    r[p.first+d[k][0]][p.second+d[k][1]] = r[p.first][p.second]+1;
                    q.push({p.first+d[k][0], p.second+d[k][1]});
                }
            }
        }
    }
    for (int i = 0; i < t; i++){
        cin >> a >> b >> c;
        a--;
        b--;
        if (c <= r[a][b] || r[a][b] == -1) cout << g[a][b] << "\n";
        else{
            c -= r[a][b];
            c %= 2;
            if (c) cout << (g[a][b]^1) << "\n";
            else cout << g[a][b] << "\n";
        }
    }
}