Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14530		Accepted: 7178		Special Judge

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. 

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.

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.

1

103000509

002109400

000704000

300502006

060000050

700803004

000401000

009205800

804000107

143628579

572139468

986754231

391542786

468917352

725863914

237481695

619275843

854396127

 #include<iostream>

 #include<cstdio>

 #include<cstring>

 #define MAXN 12

 using namespace std;

 int num[MAXN][MAXN];

 int row[MAXN][MAXN];//用来标记每一行九个数是否已被使用，使用记为1

 int col[MAXN][MAXN];//列

 int mar[MAXN][MAXN];//用0-8标识9个小数独矩阵

 char chr[MAXN][MAXN];

 bool flag=false;

 void dfs(int m,int n)

 {

     int s,i,j;

     if(m==&&n==)

     {

         for(i=;i<=;i++)

         {

             for(j=;j<=;j++)

             {

                 cout<<num[i][j];

             }

             cout<<endl;

         }

         flag=true;

     }

     else

     {

         if(num[m][n]!=)

         {

             if(n<=)

                 dfs(m,n+);

             else

                 dfs(m+,);

         }

         else

         {

             for(i=;i<=;i++)

         {

             if(row[m][i]==&&col[n][i]==&&mar[m/*+n/][i]==)//根据i,j标定它所属的行，列，和所在的小矩阵

             {

                 s=m/*+n/;

                 num[m][n]=i;

                 row[m][i]=;

                 col[n][i]=;

                 mar[s][i]=;

                 if(n<=)

                     dfs(m,n+);

                 else

                     dfs(m+,);

                 if(flag)

                     return ;

                 num[m][n]=;//一旦回溯过来，必须重新置0，因为说明此数字对于后面的搜索产生了不满足，故将其置0.

                 row[m][i]=;

                 col[n][i]=;

                 mar[s][i]=;

             }

         }

         }

     }

 }

 int main()

 {

     //freopen("data.in","r",stdin);

    // freopen("data.out","w",stdout);

     int i,j,s,t;

     cin>>t;

     while(t--)

     {

         memset(row,,sizeof(row));

         memset(col,,sizeof(col));

         memset(mar,,sizeof(mar));

         for(i=;i<;i++)

             scanf("%s",chr[i]);

         for(i=;i<;i++)

             for(j=;j<;j++)

         {

             num[i][j]=chr[i][j]-'';

             s=num[i][j];

             if(s!=)

             {

                 row[i][s]=;

                 col[j][s]=;

                 mar[i/*+j/][s]=;//初始，一旦有数字，则要将它所在的行列及小矩阵标为1

             }

         }

         flag=false;

         dfs(,);

     }

     return ;

 }