POJ 2676 Sudoku (数独 DFS)

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14368		Accepted: 7102		Special Judge

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

Sample Output

Source

Southeastern Europe 2005

题目链接：http://poj.org/problem?id=2676

http://bailian.openjudge.cn/practice/2982/

题目大意：就是数独咯，让你求解数独游戏，9乘9的矩阵要求每行每列和9个3*3的子矩阵内都出现数字1-9

题目分析：9乘9的矩阵，从第一个位置一直搜到最后一个位置，若当前位置有数字搜下一位置，否则枚举1-9并判断，

判断时当前行r = n/9当前列为c = n%9当前子矩阵的第一个元素位置为r / 3 * 3，c / 3 * 3

 1 #include <cstdio>  
 2 char s[10];  
 3 int num[9][9];  
 4 bool flag;  
 5   
 6 bool ok(int n, int cur)   
 7 {  
 8     int r = n / 9;      //当前行  
 9     int c = n % 9;      //当前列  
10     for(int j = 0; j < 9; j++)  //枚举列  
11         if (num[r][j] == cur)   
12             return false;  
13     for(int i = 0; i < 9; i++)  //枚举行  
14         if (num[i][c] == cur)   
15             return false;  
16     //得到当前所在的子矩阵的第一个元素位置  
17     int x = r / 3 * 3;    
18     int y = c / 3 * 3;  
19     //枚举子矩阵中的元素  
20     for(int i = x; i < x + 3; i++)  
21         for(int j = y; j < y + 3; j++)  
22             if (num[i][j] == cur)   
23                 return false;  
24     return true;  
25 }  
26   
27 void DFS(int n)  
28 {  
29     if(n > 80 || flag)   
30     {  
31         flag = true;  
32         return;  
33     }  
34     if(num[n / 9][n % 9])//当前位置有数字直接搜索下一位  
35     {  
36         DFS(n + 1);  
37         if(flag)  
38             return;  
39     }  
40     else  
41     {  
42         for(int cur = 1; cur <= 9; cur++)  //枚举数字  
43         {  
44             if(ok(n, cur)) //若ok则插入  
45             {  
46                 num[n / 9][n % 9] = cur;   
47                 DFS(n + 1);  
48                 if(flag)   
49                     return;  
50                 num[n / 9][n % 9] = 0; //还原  
51             }  
52         }  
53     }  
54 }  
55   
56 int main()  
57 {     
58     int T;  
59     scanf("%d", &T);  
60     while(T--)  
61     {  
62         flag = false;  
63         for(int i = 0; i < 9; i++) //得到数独矩阵  
64         {  
65             scanf("%s", s);  
66             for(int j = 0; j < 9; j++)  
67                 num[i][j] = (s[j] - '0');  
68         }  
69         DFS(0);  //从第一位开始搜  
70         for(int i = 0; i < 9; i++)  
71         {  
72             for(int j = 0; j < 9; j++)  
73                 printf("%d", num[i][j]);  
74             printf("
");        
75         }  
76     }  
77 }

View Code

题解来源：http://blog.csdn.net/tc_to_top/article/details/43699047