我的思路是利用一個二維陣列作為輔助（初始化為1），如果某一行或某一列有超過三個連續的同顏色棋子，我則將輔助陣列中的某一位置設定為0，最後再來根據輔助陣列中的0來將原陣列中的元素設定為0.

#include <bits/stdc++.h>
using namespace std;

int main() {
    int i, j, k, n, m;
    cin >> n >> m;
    
    int board[n][m];
    int bak[n][m];
    for (i = 0; i < n; ++i) {
        for (j = 0; j < m; ++j) {
            cin >> board[i][j];
            bak[i][j] = 1;
        }
    }
    for (i = 0; i < n; ++i) {
        for (j = 0; j < m; ++j) {
            if (((i + 2) < n) && (board[i+1][j] == board[i][j]) && (board[i+2][j] == board[i][j])) {
                for (k = 0; k < 3; ++k) {
                    bak[i+k][j] = 0;
                }
            }
            if (((j + 2) < m) && (board[i][j+1] == board[i][j]) && (board[i][j+2] == board[i][j])) {
                for (k = 0; k < 3; ++k) {
                    bak[i][j+k] = 0;
                }
            }
        }
    }

    for (i = 0; i < n; ++i) {
        for (j = 0; j < m; ++j) {
            if (bak[i][j] == 0) {
                board[i][j] = 0;
            }
            cout << board[i][j] << " ";
        }
        cout << endl;
    }
    cout << endl;
    return 0;
}