Sudoku

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 2372 Accepted Submission(s): 804

Problem Description

Yi Sima was one of the best counselors of Cao Cao. He likes to play a funny game himself. It looks like the modern Sudoku, but smaller.

Actually, Yi Sima was playing it different. First of all, he tried to generate a 4×4 board with every row contains 1 to 4, every column contains 1 to 4. Also he made sure that if we cut the board into four 2×2 pieces, every piece contains 1 to 4.

Then, he removed several numbers from the board and gave it to another guy to recover it. As other counselors are not as smart as Yi Sima, Yi Sima always made sure that the board only has one way to recover.

Actually, you are seeing this because you've passed through to the Three-Kingdom Age. You can recover the board to make Yi Sima happy and be promoted. Go and do it!!!

Input

The first line of the input gives the number of test cases, T(1≤T≤100). T test cases follow. Each test case starts with an empty line followed by 4 lines. Each line consist of 4 characters. Each character represents the number in the corresponding cell (one of '1', '2', '3', '4'). '*' represents that number was removed by Yi Sima.

It's guaranteed that there will be exactly one way to recover the board.

Output

For each test case, output one line containing Case #x:, where x is the test case number (starting from 1). Then output 4 lines with 4 characters each. indicate the recovered board.

Sample Input

****

2341

4123

3214

*243

*312

*421

*134

*41*

**3*

2*41

4*2*

Sample Output

Case #1:

1432

2341

4123

3214

Case #2:

1243

4312

3421

2134

Case #3:

3412

1234

2341

4123

题意：

4*4的数独游戏，最终要凑成每行每列每个分块的值为1+2+3+4；即1-4每行每列每块不能重复出现。

思路：

用DFS做，将已经填的空数为判断标记，都填好即输出并返回。列和行比较好判断，至于分块，我定义了一个计算的方式：t=row*2/2+col(row和col都是从0开始的，t表示为第几个分块)。这样很明显，从左上角的4个小分块到右下角的4个小分块分别对应的就是0-3。

代码：

 #include <cstdio>

 #include <vector>

 #include <cstring>

 using namespace std;

 struct node{

     int r,c;

 };

 vector<node>v;

 char chess[][];

 int col[][];//col[i][x]标记第i列 x数是否已经用过

 int row[][];//row[i][x]标记第i行 x数是否已经用过

 int block[][];//block[i][x]标记第i个分块，x数是否已经用过

 int ok;

 void dfs(int num){

     if(ok)  return ;

     if(num==v.size()){//所有数都填好

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

             puts(chess[i]);

         }

         ok=;

         return ;

     }

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

         int r=v[num].r;

         int c=v[num].c;

         if(!col[c][i] && !row[r][i] && !block[r/*+c/][i]){

             chess[r][c]=i+'';

             col[c][i]=row[r][i]=;

             block[r/*+c/][i]=;

             dfs(num+);

             col[c][i]=row[r][i]=;//回溯还原状态

             block[r/*+c/][i]=;

             chess[r][c]='*';

         }

     }

 }

 int main(){

     int t;

     scanf("%d",&t);

     for (int i=; i<=t; i++) {

         printf("Case #%d:\n",i);

         v.clear();

         memset(block, , sizeof(block));//初始值都为0，即都未用过

         memset(col, , sizeof(col));

         memset(row, , sizeof(row));

         ok=;

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

             scanf("%s",chess[j]);

             for (int k=; k<; k++) {

                 if(chess[j][k]=='*'){//将需要填空的点放入vector容器

                     node x;

                     x.r=j;x.c=k;

                     v.push_back(x);

                 }else{

                     int x=chess[j][k]-'';//标记已经用过的点

                     block[j/*+k/][x]=;

                     row[j][x]=;

                     col[k][x]=;

                 }

             }

         }

         dfs();//0表示在填空v[0]点

     }

     return ;

 }

巴特西

HDU 5547 Sudoku(DFS)

Sudoku

最新文章

热门文章