HDOJ 5547 Sudoku(DFS数独填数+遡及)

Sudoku
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65535/65535 K (Java/Others) Total Submission(s): 179    Accepted Submission(s): 67
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
   
   
   
   

 
4*4の行列に数を記入します.  つまり、私たちがよく言う数独の考え方です.この問題の1つの穴は、この数独が私たちの生活の中で遊んでいる数独とは異なり、斜め対角線の数は同じでも、隅々の4つの数字は同じではなく、他の純粋な暴力であればいいのです.
acコード:

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 10
#define MOD 1000000007
#define LL long long
#define INF 0xfffffff
using namespace std;
char map[MAXN][MAXN];
int check(int a,int b,char c)
{
	int i;
	for(i=0;i<4;i++)
	{
		if(map[a][i]==c||map[i][b]==c)
		return 1;
    }
    if(a/2==0&&b/2==0)
    {
    	if(map[0][0]==c||map[0][1]==c||map[1][0]==c||map[1][1]==c)
    	return 1;
	}
	else if(a/2==0&&b/2)
	{
		if(map[0][2]==c||map[0][3]==c||map[1][2]==c||map[1][3]==c)
		return 1;
	}
	else if(a/2&&b/2==0)
	{
		if(map[2][0]==c||map[2][1]==c||map[3][0]==c||map[3][1]==c)
    	return 1;
	}
	else if(a/2&&b/2)
	{
		if(map[2][2]==c||map[2][3]==c||map[3][2]==c||map[3][3]==c)
		return 1;
	}
	return 0;
}
void dfs(int x,int y)
{
	int i,j;
	if(x==4)
	{
		for(i=0;i<4;i++)
		{
			for(j=0;j<4;j++)
			printf("%c",map[i][j]);
			printf("
");
		}
		return;
	}
	if(map[x][y]!='*')
	{
		if(y==3)
		dfs(x+1,0);
		else
	    dfs(x,y+1);
    }
	else
	{
		for(i=1;i<=4;i++)
		{
			if(check(x,y,i+'0')==0)
			{
				map[x][y]=i+'0';
				if(y==3)
				dfs(x+1,0);
				else
				dfs(x,y+1);
				map[x][y]='*';//  
			}
		}
		
	}
}
int main()
{
	int t,i;
	int cas=0;
	scanf("%d",&t);
	while(t--)
	{
		for(i=0;i<4;i++)
		scanf("%s",map[i]);
		printf("Case #%d:
",++cas);
		dfs(0,0);
	}
    return 0;
}