# CDOJ 1222-Sudoku【DFS &amp;&amp; 数独】

2016-11-30 10:12:32来源:网络收集作者:Mark人点击

Sudoku
Time Limit: 3000/1000MS (Java/Others)Memory Limit: 65535/65535KB (Java/Others)

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 a4×4board
with every row contains1to4,
every column contains1to4.
Also he made sure that if we cut the board into four2×2pieces,
every piece contains1to4.

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).Ttest
cases follow. Each test case starts with an empty line followed by4lines.
Each line consist of4characters.
Each character represents the number in the corresponding cell (one of1,2,3,4).*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 containingCase
#x:, wherexis
the test case number (starting from1).
Then output4lines
with4characters
each. indicate the recovered board.

Sample input and output

Sample Input

Sample Output
3
****
2341
4123
3214
*243
*312
*421
*134
*41*
**3*
2*41
4*2*Case #1:
1432
2341
4123
3214
Case #2:
1243
4312
3421
2134
Case #3:
3412
1234
2341
4123Source

The 2015 China Collegiate Programming Contest

#include
#include
#include
using namespace std;
int d;
struct node{
int x, y;
};
node str[50];
bool vish[5][5];
bool visl[5][5];
bool visk[5][5];
char map[5][5];
int change (int x, int y){
return x /2 * 2 + y / 2;
}
void dfs(int k){
if(k == d){
for(int i = 0; i < 4; ++i){
printf("%s/n", map[i]);
}
return ;
}
for(int i = 1; i <= 4; ++i){
int t = change(str[k].x, str[k].y);
if(!vish[str[k].x][i] && !visl[str[k].y][i] && !visk[t][i]){
map[str[k].x][str[k].y] = i + '0';
vish[str[k].x][i] = 1;
visl[str[k].y][i] = 1;
visk[t][i] = 1;
dfs(k + 1);
vish[str[k].x][i] = 0;
visl[str[k].y][i] = 0;
visk[t][i] = 0;
map[str[k].x][str[k].y] = '*';
}
}
}
int kcase;
int main (){
int T;
kcase = 1;
scanf("%d", &T);
while(T--){
d = 0;
memset(vish, 0, sizeof(vish));
memset(visl, 0, sizeof(visl));
memset(visk, 0, sizeof(visk));
for(int i = 0; i < 4; ++i){
scanf("%s", map[i]);
for(int j = 0; j < 4; ++j)
if(map[i][j] != '*'){
vish[i][map[i][j] - '0'] = 1;
visl[j][map[i][j] - '0'] = 1;
visk[change(i, j)][map[i][j] - '0'] = 1;
}
else {
str[d].x = i;
str[d].y = j;
d++;
}
}
printf("Case #%d:/n", kcase++);
dfs(0);
}
return 0;
}