[Ybtoj Chapter 4 Example 2] Sudoku game [Deep Search]

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Problem solving ideas

This problem is realized by searching.
We can use three arrays to mark whether the row, column and Jiugongge numbers have been filled in.
Then simulate the digital filling process and consider all one to nine at once.

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Code

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;

int a[10][10],kj[10][10],lyx[10][10],bf[10][10][10];
bool flag;
string s;

int check(int x)//九宫格编号
{
    
    
    if(x>=1&&x<=3)
        return 1;
    if(x>3&&x<=6)
        return 2;
    if(x>6&&x<=9)
        return 3;
}

void dfs(int x,int y)
{
    
    
	if(flag==1)	
		return;
    if(x>9&&y==1)
    {
    
    
        for(int i=1; i<=9; i++)
        	for(int j=1; j<=9; j++)
            	printf("%d",a[i][j]);
        printf("\n");
        flag=1;
        return;
    }
    if(a[x][y]!=0)//如果已有固定的数
    {
    
    
    	if(y==9)
    		dfs(x+1,1);
    	else dfs(x,y+1);
    	return;
	}
	for(int i=1;i<=9;i++)
	{
    
    
		if(!kj[x][i]&&!lyx[y][i]&&!bf[check(x)][check(y)][i])//尝试填数
		{
    
    
			a[x][y]=i;
			kj[x][i]=1;
			lyx[y][i]=1;
			bf[check(x)][check(y)][i]=1;
			if(y==9)dfs(x+1,1);
    		else dfs(x,y+1);
    		a[x][y]=0;
			kj[x][i]=0;
			lyx[y][i]=0;
			bf[check(x)][check(y)][i]=0;
		}
	}
}

int main()
{
    
    
    while(cin>>s)
    {
    
    
    	memset(a,0,sizeof(a));
    	memset(kj,0,sizeof(kj));
    	memset(lyx,0,sizeof(lyx));
    	memset(bf,0,sizeof(bf));
    	flag=0;
        if(s=="end")
            break;
        for(int i=1; i<=9; i++)
        {
    
    
            for(int j=1; j<=9; j++)
            {
    
    
                if(s[(i-1)*9+j-1]>='1'&&s[(i-1)*9+j-1]<='9')
                {
    
    
                    a[i][j]=s[(i-1)*9+j-1]-48;
                    kj[i][a[i][j]]=1;
                    lyx[j][a[i][j]]=1;
                    bf[check(i)][check(j)][a[i][j]]=1;//记录行、列、九宫格
                }
            }
        }
		dfs(1,1);
    }
}