E - Sudoku

Sudoku求解算法

最新推荐文章于 2020-12-22 12:00:52 发布

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E - Sudoku

Time Limit:2000MS Memory Limit:65536KB 64bit IO Format:%lld & %llu

Submit Status

Description

Sudoku is a very simple task. A square table with 9 rows and 9 columns is divided to 9 smaller squares 3x3 as shown on the Figure. In some of the cells are written decimal digits from 1 to 9. The other cells are empty. The goal is to fill the empty cells with decimal digits from 1 to 9, one digit per cell, in such way that in each row, in each column and in each marked 3x3 subsquare, all the digits from 1 to 9 to appear. Write a program to solve a given Sudoku-task.

Input

The input data will start with the number of the test cases. For each test case, 9 lines follow, corresponding to the rows of the table. On each line a string of exactly 9 decimal digits is given, corresponding to the cells in this line. If a cell is empty it is represented by 0.

Output

For each test case your program should print the solution in the same format as the input data. The empty cells have to be filled according to the rules. If solutions is not unique, then the program may print any one of them.

Sample Input

Sample Output

book[i][j][k] i=0时，是按行标记，i=1时，是按列标记，i==2时，是按方块标记，j是指要标记的那一行（或列或方块），k是指k这个数出没出现过，遍历一遍，将所有0的坐标存起来，然后对所有零所在点进行搜索；

#include<stdio.h>
#include<string.h>
int book[4][12][12],a[12][12],n,zero,b[12][12],flog=0;
struct node
{
	int x,y;
}index[200];
int dfs(int m);
int main()
{
	int i,j;
	//for(i=0;i<9;i++)
//	{
	//	for(j=0;j<9;j++)
	//	printf("%d ",(i/3)*3+(j/3)+1);
	//	printf("\n");
//	}
	char ch[12][12];
	scanf("%d",&n);
	getchar();
	while(n--)
	{
		flog=0;
		zero=0;
		memset(book,0,sizeof(book));
		for(i=0;i<9;i++)
		scanf("%s",ch[i]);
		for(i=0;i<9;i++)
		for(j=0;j<9;j++)
		{
			a[i][j]=ch[i][j]-'0';
			if(a[i][j]==0)
			{
				index[zero].x=i;
				index[zero].y=j;
				zero++;
			}
			else
			{
				b[i][j]=a[i][j];
				book[0][i][a[i][j]]=1;
				book[1][j][a[i][j]]=1;
				book[2][(i/3)*3+(j/3)+1][a[i][j]]=1;
			}
		}
	//	for(int k=0;k<3;k++)
	//	for(i=0;i<9;i++)
	//	{
	//		for(j=1;j<=9;j++)
	//		printf("%d ",book[k][i][j]);
	//		printf("\n");
	//	}
		dfs(0);
		for(i=0;i<9;i++)
		{
			for(j=0;j<9;j++)
			printf("%d",b[i][j]);
			printf("\n");
		}
	}
	return 0;
}
int dfs(int m)
{
	int tx,ty;
	if(m==zero||flog)
	{
		flog=1;
		return 0;
	}
	tx=index[m].x,ty=index[m].y;
	for(int i=1;i<=9;i++)
	{
		if( book[0][tx][i]==0&&book[1][ty][i]==0&&book[2][(tx/3)*3+(ty/3)+1][i]==0)
			{
			//	for(int k=0;k<9;k++)
	//	{
		//	for(int j=0;j<9;j++)
		//	printf("%d",b[k][j]);
		//	printf("\n");
		//}
	//	printf("\n");
				book[0][tx][i]=1;
				book[1][ty][i]=1;
				book[2][(tx/3)*3+(ty/3)+1][i]=1;
				b[tx][ty]=i;
				dfs(m+1);
				if(flog)
				break;
				b[tx][ty]=1;
				book[0][tx][i]=0;
				book[1][ty][i]=0;
				book[2][(tx/3)*3+(ty/3)+1][i]=0;
			}
	}
	return 0;
}