--- /dev/null
+/*
+ * A new method to solve the puzzles,
+ * more nice orgnized, and more efficent I suppose.
+ * (at least compared to my original one)
+ *
+ * this one take puzzles and std input,
+ * an example puzzle could be found below:
+ * 0 0 0 2 6 0 7 0 1
+ * 6 8 0 0 7 0 0 9 0
+ * 1 9 0 0 0 4 5 0 0
+ * 8 2 0 1 0 0 0 4 0
+ * 0 0 4 6 0 2 9 0 0
+ * 0 5 0 0 0 3 0 2 8
+ * 0 0 9 3 0 0 0 7 4
+ * 0 4 0 0 5 0 0 3 6
+ * 7 0 3 0 1 8 0 0 0
+ *
+ * after calculation, the correct output should be:
+ * +------+------+-----
+ * |4 3 5 |2 6 9 |7 8 1
+ * |6 8 2 |5 7 1 |4 9 3
+ * |1 9 7 |8 3 4 |5 6 2
+ * +------+------+-----
+ * |8 2 6 |1 9 5 |3 4 7
+ * |3 7 4 |6 8 2 |9 1 5
+ * |9 5 1 |7 4 3 |6 2 8
+ * +------+------+-----
+ * |5 1 9 |3 2 6 |8 7 4
+ * |2 4 8 |9 5 7 |1 3 6
+ * |7 6 3 |4 1 8 |2 5 9
+ *
+ * Yet, the core computing part is still a kind of brute force,
+ * which is more intutive to implement.
+ * It might be implemented to my CheaterHub project in the future,
+ * but I'm getting really frustrated by 81 boxes in UI design.
+ */
+#include <iostream>
+#include <vector>
+
+using namespace std;
+
+int main()
+{
+ vector<vector<int>> ori(9, vector<int>(9));
+ vector<pair<int, int>> todo;
+ int cur = 0;
+
+ // Another idea, is pre-generate checking list
+ // for each blank box in the puzzle,
+ // it should save some runtime for less checking.
+ for(int i = 0; i < 9; i++)
+ {
+ for(int j = 0; j < 9; j++)
+ {
+ cin >> ori[i][j];
+ if(!ori[i][j])
+ todo.push_back({i, j});
+ }
+ }
+
+ cout << "********************" << endl;
+
+ while(cur < todo.size())
+ {
+ int x = todo[cur].first;
+ int y = todo[cur].second;
+ vector<bool> check(10, false);
+
+ // TODO: there are 6 overlapping checks,
+ // including (x, y) itself twice.
+ // column
+ for(int i = 0; i < 9; i++)
+ check[ori[i][y]] = true;
+ // row
+ for(int j = 0; j < 9; j++)
+ check[ori[x][j]] = true;
+ // 9x9 box
+ int ibase = x / 3 * 3;
+ int jbase = y / 3 * 3;
+ for(int i = 0; i < 3; i++)
+ for(int j = 0; j < 3; j++)
+ check[ori[i+ibase][j+jbase]] = true;
+
+ int k = ori[x][y] + 1;
+ for(; k < 10; k++)
+ if(!check[k]) break;
+
+ if(k > 9)
+ {
+ ori[x][y] = 0;
+ cur--;
+ continue;
+ }
+ else
+ {
+ ori[x][y] = k;
+ cur++;
+ }
+ }
+
+ for(int i = 0; i < 9; i++)
+ {
+ if(i%3 == 0) cout << "+------+------+-----" << endl;
+ for(int j = 0; j < 9; j++)
+ {
+ if(j%3 == 0) cout << "|";
+ cout << ori[i][j] << " ";
+ }
+ cout << endl;
+ }
+
+ return 0;
+}