题解 | #本场比赛灵感来源于树状数组出题组#

这条题目用的偏移量枚举，这种构造题多举几个例子其实挺简单的（一开始以为跟bfs dfs有关）

using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n, m, q;
    cin >> n >> m >> q;
    vector<vector<long long>> grid(n + 2, vector<long long>(m + 2, 0));
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> grid[i][j];
        }
    }
    vector<pair<int, int>> dirs = {
        {-2, 0}, {-1, -1}, {-1, 0}, {-1, 1},
        {0, -2}, {0, -1}, {0, 0}, {0, 1}, {0, 2},
        {1, -1}, {1, 0}, {1, 1}, {2, 0}
    };

    vector<vector<long long>> val(n + 2, vector<long long>(m + 2, 0));
    for (int x0 = 1; x0 <= n; ++x0) {
        for (int y0 = 1; y0 <= m; ++y0) {
            for (auto& d : dirs) {
                int dx = d.first;
                int dy = d.second;
                int x = x0 + dx;
                int y = y0 + dy;
                if (x >= 1 && x <= n && y >= 1 && y <= m) {
                    val[x0][y0] += grid[x][y];
                }
            }
        }
    }
    long long max_val = -1;
    int best_x = 1, best_y = 1;
    for (int x = 1; x <= n; ++x) {
        for (int y = 1; y <= m; ++y) {
            if (val[x][y] > max_val) {
                max_val = val[x][y];
                best_x = x;
                best_y = y;
            }
        }
    }
    while (q--) {
        int x, y, z;
        cin >> x >> y >> z;
        for (auto& d : dirs) {
            int dx = d.first;
            int dy = d.second;
            int x0 = x + dx;
            int y0 = y + dy;
            if (x0 >= 1 && x0 <= n && y0 >= 1 && y0 <= m) {
                val[x0][y0] += z;
                if (val[x0][y0] > max_val) {
                    max_val = val[x0][y0];
                    best_x = x0;
                    best_y = y0;
                }
            }
        }

        cout << best_x << " " << best_y << "\n";
    }

    return 0;
}