TOJ 3287 Sudoku dfs

忘是亡心i 2022-05-19 14:08 150阅读 0赞

## **3287: Sudoku** ##

**描述**

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.  
![3287.jpg][]

**输入**

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.

**输出**

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.

**样例输入**

1  
103000509  
002109400  
000704000  
300502006  
060000050  
700803004  
000401000  
009205800  
804000107

**样例输出**

143628579  
572139468  
986754231  
391542786  
468917352  
725863914  
237481695  
619275843  
854396127

数独游戏，每个格子填写1-9的数字，要求3\*3的小格子里不能有重复数字，9\*9的大格子里每行每列不能有重复的数字。所以要check三下，分别是行，列和3\*3的小方阵。（小方阵的序号以3\*(i/3)+j/3来表示）

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    int a[20][20],flag;
    int r[20][20],l[20][20],s[20][20];//r行 l列 s小方阵 
    void dfs(int i,int j)
    {
    	if(i==9&&j==0)
    	flag=1;	
    	else
    	{
    		if(a[i][j]!=0)
    		{
    			if(j<8)
    			dfs(i,j+1);
    			else
    			dfs(i+1,0);
    		}
    		else
    		{
    			for(int k=1;k<=9;k++)
    			{
    				if(!r[i][k]&&!l[j][k]&&!s[3*(i/3)+j/3][k])
    				{
    					a[i][j]=k;
    					r[i][k]=1;
    					l[j][k]=1;
    					s[3*(i/3)+j/3][k]=1;
    					if(j<8)
    					dfs(i,j+1);
    					else
    					dfs(i+1,0);
    					if(flag)return;
    					//回溯重置 
    					a[i][j]=0;
    					r[i][k]=0;
    					l[j][k]=0;
    					s[3*(i/3)+j/3][k]=0;
    				}
    			}
    		}
    	}
    }
    int main()
    {
    	int t,i,j;
    	scanf("%d",&t);
    	while(t--)
    	{
    		memset(r,0,sizeof r);
    		memset(l,0,sizeof l);		
    		memset(s,0,sizeof s);
    		for(i=0;i<9;i++)
    		{
    			for(j=0;j<9;j++)
    			{
    				scanf("%1d",&a[i][j]);
    				if(a[i][j])
    				{
    					r[i][a[i][j]]=1;
    					l[j][a[i][j]]=1;
    					s[3*(i/3)+j/3][a[i][j]]=1;
    				}
    			}
    		}
    		flag=0;
    		dfs(0,0);
    		for(i=0;i<9;i++)
    		{
    			for(j=0;j<9;j++)
    			{
    				printf("%d",a[i][j]);
    			}
    			printf("\n");
    		}
    	}
    }

[3287.jpg]: http://210.32.82.1/acmhome/judge/images/3287.jpg