POJ-2676-Sudoku

Sudoku

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17913		Accepted: 8674		Special Judge

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.
输出

为每个测试用例的程序应该在相同的格式打印解决方案作为输入数据。 空的细胞必须按照规定。 如果解决方案并不是唯一的,程序可以打印任何其中之一。
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

#include<iostream>
#include<string.h>
#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std;
int a[10][10], flag;
bool vis1[10],vis2[10],vis3[10];
bool check(int x, int y)
{
memset(vis1,0,sizeof(vis1));
memset(vis2,0,sizeof(vis2));
memset(vis3,0,sizeof(vis3));
for(int i = 1;i<=9;++i)
{
if(a[i][y]&&vis1[a[i][y]])return false;
else vis1[a[i][y]] = 1;
if(a[x][i]&&vis2[a[x][i]])return false;
else vis2[a[x][i]] = 1;
}
int aa, b;
if(x<=3)aa = 1;
else if(x<=6)aa = 4;
else aa = 7;
if(y<=3)b = 1;
else if(y<=6)b = 4;
else b = 7;
for(int i = aa;i<=aa+2;++i)
{
for(int j = b;j<=b+2;++j)
{
if(a[i][j]&&vis3[a[i][j]])
return false;
else vis3[a[i][j]] = 1;
}
}
return 1;
}
void dfs(int x, int y)
{
if(flag)return;
if(x>9)
{
x = 1;
y++;
}
if(x==1&&y==10)
{
flag = 1;
return;
}
if(a[x][y])dfs(x+1, y);
else
{
for(int i =1; i<=9;++i)
{
a[x][y] = i;
if(check(x,y))dfs(x+1, y);
if(flag)return;
}
a[x][y] = 0;
}
}
int main()
{
int t;
scanf("%d", &t);
while(t--)
{
flag = 0;
for(int i = 1;i<=9;++i)
for(int j = 1;j<=9;++j)
scanf("%1d",&a[i][j]);
dfs(1, 1);
for(int i = 1;i<=9;++i)
{
for(int j = 1;j<=9;++j)
printf("%d", a[i][j]);
printf("\n");
}
}
return 0;
}