Sudoku Sudoku decisive dfs (), each box is 3 * 3 to count when determining the position of the upper left corner of the box ...

Problem 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

1

103000509

002109400

000704000

300502006

060000050

700803004

000401000

009205800

804000107

 

 

Sample Output

143628579

572139468

986754231

391542786

468917352

725863914

237481695

619275843

854396127

***************************************************************************************************************************

3 * 3 grid calculated upper-left corner of {(i / 3) * 3, (j / 3) * 3}

***************************************************************************************************************************

 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cstdio>
 6 using namespace std;
 7 int num,su[1001][2],map[11][11],cas;
 8 int i,j,k;
 9 bool judge(int x,int y,int k)
10  {
11     int it,jt;
12     //同行同列时的判断
13     for(it=0;it<9;it++)
14     {
15         if(map[it][y]==k)return false;
16         if(map[x][it]==k)return false;
17     }
18     it=(x/3)*3;//小方格左上角的数
19     jt=(y/3)*3;
20     for(int st=0;st<3;st++)
21      for(intYT = 0 ; YT < . 3 ; YT ++ )
 22 is         IF (Map [IT + ST] [YT + JT] == K)
 23 is          return  to false ;
 24      return  to true ;
 25  }
 26 is  int DFS ( int CAP)
 27  {
 28      int IE ;
 29      IF (CAP < 0 ) return  . 1 ; // real conditions of 
30      for (IE = . 1 ; IE <= . 9 ; ++ IE )
 31 is      {
 32          intXS = SU [CAP] [ 0 ];
 33 is          int YS = SU [CAP] [ . 1 ];
 34 is          IF (Judge (XS, YS, IE))
 35          {
 36              Map [XS] [YS] = IE;
 37 [              IF ( DFS (CAP- . 1 )) return  . 1 ; // search in the end, the condition returns 
38 is              Map [XS] [YS] = 0 ; // restore 
39          }
 40      }
 41 is      return  0 ;
 42 is  }
 43 is  int main ()
 44 is  {
 45     scanf("%d",&cas);
46     getchar();
47     while(cas--)
48     {
49         char ch;
50         num=0;
51         for(i=0;i<9;i++,getchar())
52         {
53             for(j=0;j<9;j++)
54             {
55                 scanf("%c",&ch);
56                 Map [I] [J] = CH-, ' 0 ' ;
 57 is                  IF (Map [I] [J] == 0 ) // record the number of positions to be filled 
58                  {
 59                      SU [NUM] [ 0 ] = I ;
 60                      SU [NUM ++] [ . 1 ] = J;
 61 is                  }
 62 is              }
 63 is          }
 64          DFS (num- . 1 );
 65          // output 
66          for (I = 0 ; I < . 9 ; I ++ )
 67         {
68          for(j=0;j<9;j++)
69            printf("%d",map[i][j]);
70          printf("\n");
71         }
72         printf("\n");
73 
74     }
75     return 0;
76 
77 }

View Code

 

 

Reproduced in: https: //www.cnblogs.com/sdau--codeants/p/3372893.html