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 4×44×4 board with every row contains 11 to 44 , every column contains 11 to 44 .
Also he made sure that if we cut the board into four 2×22×2
pieces, every piece contains 11
to 44
.
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, TT
(1≤T≤1001≤T≤100
). TT
test cases follow. Each test case starts with an empty line followed by 44
lines.
Each line consist of 44
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 xx
is the test case number (starting from 11
).
Then output 44
lines with 44
characters each. indicate the recovered board.
Sample Input
3 **** 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：一个4*4的矩阵，满足每个数的同行同列都由1234构成的，把4*4矩阵分成4个2*2矩阵，每个矩阵的四个元素也分别为1234。这题就是暴力深搜。ac代码：
1 #include<cstdio>
2 #include<cstring>
3 #define Max 8
4 char st[Max][Max];
5 int judg(int x,int y,char i) //判断当前位置可以是什么数字
6 {
7 if(x<3&&y<3)
8 {
9 if(st[1][1]==i||st[1][2]==i||st[2][1]==i||st[2][2]==i)
10 return 0;
11 }
12 else if(x<3&&y>2)
13 {
14 if(st[1][3]==i||st[1][4]==i||st[2][3]==i||st[2][4]==i)
15 return 0;
16 }
17 else if(x>2&&y<3)
18 {
19 if(st[3][1]==i||st[3][2]==i||st[4][1]==i||st[4][2]==i)
20 return 0;
21 }
22 else if(x>2&&y>2)
23 {
24 if(st[3][3]==i||st[3][4]==i||st[4][3]==i||st[4][4]==i)
25 return 0;
26 }
27 for(int k=1;k<=4;k++)
28 {
29 if(st[x][k]==i||st[k][y]==i)
30 return 0;
31 }
32 return 1;
33 }
34 void dfs(int x,int y)
35 {
36 if(x==5)
37 {
38 for(int i=1;i<=4;i++)
39 printf("%s\n",st[i]+1);
40 return ;
41 }
42 if(st[x][y]!=‘*‘)
43 {
44 if(y==4)
45 dfs(x+1,1);
46 else
47 dfs(x,y+1);
48 }
49 else
50 {
51 for(int i=1;i<=4;i++)
52 {
53 if(judg(x,y,i+‘0‘))
54 {
55 st[x][y]=i+‘0‘;
56 if(y==4)
57 dfs(x+1,1);
58 else
59 dfs(x,y+1);
60 st[x][y]=‘*‘; //如果没有满足的则回溯
61 }
62 }
63 }
64 }
65 int main()
66 {
67 int t;
68 scanf("%d",&t);
69 getchar();
70 for(int i=1;i<=t;i++)
71 {
72 for(int j=1;j<=4;j++)
73 scanf("%s",st[j]+1);
74 printf("Case #%d:\n",i);
75 dfs(1,1);
76 }
77 return 0;
78 }