Sudoku
Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 107 Accepted Submission(s) : 34
Special Judge
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
1
103000509
002109400
000704000
300502006
060000050
700803004
000401000
009205800
804000107
Sample Output
143628579
572139468
986754231
391542786
468917352
725863914
237481695
619275843
854396127
题意：很有意思 数独
思路：之前做过数独杀手的一个，直接在源代码的基础上修改标识符和输入输出的差异 以为差不多
写死机了两次 结束都结束不掉 一直tle
找了个比较好理解的题解，对于填数据，都是采用的暴力
[i][j] 各数组分别代表当前行有没有j，当前列有么有j，每个小九宫格有没有j 这样很好理解
还有一种解法就是每行每列每个小九宫格单独判断有没有将要填上的数据，判断当前vis，没试过就一直试，
两种方式最关键的都是恢复
Source Code#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std
;
int map
[12][12];
int judgex
[12][12];
int judgey
[12][12];
int judgenine
[12][12];
int dfs(int x
,int y
)
{
int i
,j
,num
;
if(x
==10)
return 1;
int flag
=0;
if(map
[x
][y
]!=0)
{
if(y
==9)
flag
=dfs(x
+1,1);
else
flag
=dfs(x
,y
+1);
if(flag
)
return 1;
else
return 0;
}
else{
num
=3*((x
-1)/3)+(y
-1)/3+1;
for(i
=1;i
<=9;i
++)
if(!judgex
[x
][i
]&&!judgey
[y
][i
]&&!judgenine
[num
][i
])
{
map
[x
][y
]=i
;
judgex
[x
][i
]=1;
judgey
[y
][i
]=1;
judgenine
[num
][i
]=1;
if(y
==9)
flag
=dfs(x
+1,1);
else
flag
=dfs(x
,y
+1);
if(flag
)
return 1;
else {
judgex
[x
][i
]=0;
judgey
[y
][i
]=0;
judgenine
[num
][i
]=0;
map
[x
][y
]=0;
}
}
}
return 0;
}
int main()
{
int t
,i
,j
,num
,ans
;
char a
;
cin
>>t
;
while(t
--)
{
memset(map
,0,sizeof(map
));
memset(judgex
,0,sizeof(judgex
));
memset(judgey
,0,sizeof(judgey
));
memset(judgenine
,0,sizeof(judgenine
));
for(i
=1;i
<=9;i
++)
{
for(j
=1;j
<=9;j
++)
{
cin
>>a
;
map
[i
][j
]=a
-'0';
if(map
[i
][j
])
{
num
=((i
-1)/3)*3+(j
-1)/3+1;
judgex
[i
][map
[i
][j
]]=1;
judgey
[j
][map
[i
][j
]]=1;
judgenine
[num
][map
[i
][j
]]=1;
}
}
}
dfs(1,1);
for(i
=1;i
<=9;i
++)
{
for(j
=1;j
<=9;j
++)
cout
<<map
[i
][j
];
cout
<<endl
;
}
}
return 0;
}
