Sudoku 数独问题(dfs)

最新推荐文章于 2021-03-05 14:45:04 发布

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最新推荐文章于 2021-03-05 14:45:04 发布

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分类专栏： ACM- 搜索(DFS BFS)

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本文介绍了使用深度优先搜索(DFS)算法解决数独问题的方法，详细阐述了算法思路，并通过实例展示了如何通过DFS填充数独网格。

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Problem 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

#if 0
#include<iostream>
using namespace std;
int a[15][15],cnt;
bool flag=0;

int ok(int x,int y,int k)
{
	for(int i=0; i<9; i++)
	{
		if(a[x][i]==k)
			return 0;
		if(a[i][y]==k)
			return 0;	
	}
	
	int sx=x-x%3;
	int sy=y-y%3;
	for(int i=sx; i<sx+3; i++)
	{
		for(int j=sy; j<sy+3; j++)
		{
			if(a[i][j]==k)
				return 0;
		}
	}
	return 1;
}


void dfs(int num)
{
	
	bool flag1=0;
	int x,y;

	if(num==cnt)  
	{
		for(int i=0; i<9; i++)
		{
			for(int j=0; j<9; j++)
			{
				cout<<a[i][j];
			}
			cout<<endl;
		}
		flag=1;
		return;
	}
	if(flag)
		return;
    
	for(int i=0; i<9; i++)
	{
		for(int j=0; j<9; j++)
		{
			if(a[i][j]==0)
			{
				flag1=1;
				x=i;y=j;
				break;
			}
		}
		if(flag1)
			break;
	}
			
		for(int k=1; k<=9; k++)
		{
			if(ok(x,y,k))
			{
				a[x][y]=k;
				dfs(num+1);
				a[x][y]=0;
			}
		
		}		
}

int main()
{
	int N;
	char c;
	cin>>N;
	
	while(N--)
	{
		flag=0;
		cnt=0;
		
		for(int i=0; i<9; i++)
		{
			for(int j=0; j<9; j++)
			{
				cin>>c;
				a[i][j]=c-'0';
				if(a[i][j]==0)
					cnt++;	 
			}
		}
	
		dfs(0);

		
	}
		
}
#endif