hdu 5547 Sudoku

Sudoku

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 710    Accepted Submission(s): 268

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

  

   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
  

 

Source

The 2015 China Collegiate Programming Contest

 

Recommend

wange2014

 

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=5547

题目大意：4个数字的数独。

解体思路：dfs，枚举填入数字1-4，每次判断填入数字是否符合就行。同poj2676。

代码如下：

#include <cstdio>
#include <cstring>
char s[10];
int a[10][10];
bool p;
bool ok(int x,int y)
{
	for(int i=0;i<4;i++)
	{
		if(i!=x && a[i][y]==a[x][y])
			return false;
	}
	for(int i=0;i<4;i++)
	{
		if(i!=y && a[x][i]==a[x][y])
			return false;
	}
	int nx=(x/2)*2;
	int ny=(y/2)*2;
	for(int i=nx;i<nx+2;i++)
		for(int j=ny;j<ny+2;j++)
		{
			if(i==x && j==y)
				continue;
			if(a[x][y]==a[i][j])
				return false;
		}
	return true;
}
void dfs(int x,int y)
{
	if(p || x==4)
	{
		p=true;
		return ;
	}
	if(a[x][y])
	{
		if(y==3)
			dfs(x+1,0);
		else
			dfs(x,y+1);
		if(p)
			return;
	}
	else
	{
		for(int i=1;i<=4;i++)
		{
			a[x][y]=i;
			if(ok(x,y))
			{
			//	printf("%d\n",i);
				if(y==3)
					dfs(x+1,0);
				else
					dfs(x,y+1);
			}
			if(p)
				return;
			a[x][y]=0;
		}
	}
}

int main(void)
{
	int n;
	scanf("%d",&n);
	for(int ca=1;ca<=n;ca++)
	{
		p=false;
		memset(a,0,sizeof(a));
		for(int i=0;i<4;i++)
		{
			scanf("%s",s);
			for(int j=0;j<4;j++)
			{
				if(s[j]=='*')
					a[i][j]=0;
				else
 					a[i][j]=s[j]-'0';
			}
		}
		dfs(0,0);
		printf("Case #%d:\n",ca );
		for(int i=0;i<4;i++)
		{
			for(int j=0;j<4;j++)
				printf("%d",a[i][j] );
			printf("\n");
		}
	}
}