POJ2676：Sudoku（DFS）

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task. 

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

1103000509002109400000704000300502006060000050700803004000401000009205800804000107

Sample Output

143628579572139468986754231391542786468917352725863914237481695619275843854396127

前后用零散的时间想了两天，暴力DFS了半天，终于给我AC过了

我现在好兴奋♂

好了，正题，其实也就是行、列、宫里没有重复的数

我用三个布尔型的二维数组分别保存上面三种数据，

第一维表示序号，第二维保存数1~9，到这些数中的某个数被使用就是其为true

大概思路就是这样，下面说一下剪枝。

一开始我直接遍历数独的每一个格子，然后，毫无疑问肯定会TLE

后面突然想到，只需把为零的格子全部先保存，然后一个一个遍历就行了

#include <stdio.h>#include <algorithm>using namespace std;int map[10][10];bool cols[9][10], rows[9][10], grid[9][10];int sign;struct{int x, y;}zero[100];void dfs(int total){if (sign)return;if (!total){for (int m = 0; m < 9; m++){for (int n = 0; n < 9; n++)printf("%d", map[m][n]);printf("\n");}sign = 1;return;}int x = zero[total-1].x, y = zero[total-1].y;int pos = x / 3*3+ y / 3;if (map[x][y]){dfs(total);return;}for (int i = 1; i <= 9; i++){if (!rows[x][i] && !cols[y][i] && !grid[pos][i]){rows[x][i] = true;cols[y][i] = true;grid[pos][i] = true;map[x][y] = i;dfs(total-1);map[x][y] = 0;rows[x][i] = false;cols[y][i] = false;grid[pos][i] = false;}}return;}int main(){int t;char str[10];scanf("%d", &t);while (t--){int sum= 0;sign = 0;memset(cols, 0, sizeof(cols));memset(rows, 0, sizeof(rows));memset(grid, 0, sizeof(grid));for (int i = 0; i < 9; i++){scanf("%s",str);for (int j = 0; j < 9; j++){map[i][j] = str[j] - '0';if (map[i][j] != 0){rows[i][map[i][j]] = true;cols[j][map[i][j]] = true;int pos = j / 3 + i / 3 * 3;grid[pos][map[i][j]] = true;}else{zero[sum].x = i;zero[sum++].y = j;}}}dfs(sum);}return 0;}