1 题目

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。

2 解法

核心思想是, 首先把所有未确定值的位置收集起来, 让每个位置都选取其可选的值, 这就要有数据结构记录下其所出的行, 列, 以及块已经被选的值, 这样遍历其可选的值, 并且如果之前选的值不合适, 引入回退机制.

代码:

 

int row[9];
int col[9];
int block[3][3];
bool valid;
vector<pair<int, int>> blanks;
vector<vector<char>> *pBoard;void flip(int row_index, int col_index, int number, bool changeBoard) {row[row_index] ^= (1 << number);col[col_index] ^= (1 << number);block[row_index / 3][col_index / 3] ^= (1 << number);if (changeBoard) {if ((*pBoard)[row_index][col_index] == '.') {(*pBoard)[row_index][col_index] = '1' + number;} else {(*pBoard)[row_index][col_index] = '.';}}
}void dfs(int blanks_index) {if (blanks_index == blanks.size()) {valid = true;return;}int row_index = blanks[blanks_index].first;int col_index = blanks[blanks_index].second;int number = 0;while (true) {if (number == 9)return;int bin_num = (1 << number);if (row[row_index] & bin_num || col[col_index] & bin_num || block[row_index / 3][col_index / 3] & bin_num) {number ++;continue;}flip(row_index, col_index, number, true);dfs(blanks_index + 1);if (!valid) {flip(row_index, col_index, number, true);number ++;}}
}void solveSudoku(vector<vector<char>>& board) {memset(row, 0, sizeof(row));memset(col, 0, sizeof(col));memset(block, 0, sizeof(block));valid = false;pBoard = &board;for (int i = 0; i < 9; i ++) {for (int j = 0; j < 9; j ++) {if (board[i][j] == '.') {blanks.push_back(make_pair(i, j));} else {int number = board[i][j] - '1';flip(i, j, number, false);}}}dfs(0);
}