poj2676(Sudoku)

最新推荐文章于 2022-08-11 20:58:55 发布

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#each #input #output #string #table #存储

本文介绍了一种基于回溯算法的Sudoku求解方法。通过使用布尔数组记录每行、每列及每个3x3子区域已使用的数字，实现对空格的有效填充。程序能够处理并解决标准9x9 Sudoku谜题。

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/**

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

Source

Southeastern Europe 2005

**/

#include <iostream>
#define clear(array) memset(array,false,sizeof(array))
using namespace std;

bool Row[9][10]; //用于标识第i行用到了1-9中的哪些数。R[i][j] = true表示第i行数j已经使用过了
bool Column[9][10]; //用于标识第i列用到了1-9中的哪些数。说明同上
bool Block[9][10]; //用于标识每一个3*3列的方块中用到了1-9中的哪些数
//bool Tag[9][9];
int A[9][9]; //存储输入数据

void init()
{
clear(Row); //初始化数组，下同
clear(Column);
clear(Block);
//clear(Tag);
memset(A,0,sizeof(A));
}
void input()
{
int temp = 0;

for (int i=0; i<9; ++i)
{
  for (int j=0; j<9; ++j)
  {
   scanf("%1d",&temp);
   if (temp) //如果输入不为0
   {
    A[i][j] = temp;
    Row[i][temp] = true;
    Column[j][temp] = true;
    Block[3*(i/3)+j/3][temp] = true;
   }
  }
}
cout << endl;
}

bool sudoku(int n)
{
int i = n / 9; //9*9中数组的行号
int j = n % 9; //列号

if (n > 80) return true; //搜索到最后一个节点返回
if (!A[i][j])
{
  for (int k=1; k<=9; ++k)
  {
   if (!Row[i][k] && !Column[j][k] && !Block[3*(i/3)+j/3][k])//k可以放在当前位置
   {
    A[i][j] = k;
    Row[i][k] = true;
    Column[j][k] = true;
    Block[3*(i/3)+j/3][k] = true;

if (sudoku(n+1)) return true; //搜索解决下一个位置

    A[i][j] = 0;
    Row[i][k] = false;
    Column[j][k] = false;
    Block[3*(i/3)+j/3][k] = false;

   }
  }
  return false; //此处的return 很重要。切记：回溯递归时正确return很重要。一个if一个return
}
if (sudoku(n+1)) return true;

return false;
}

void output()
{
for (int i=0; i<9; ++i)
{
  for (int j=0; j<9; ++j)
  {
   cout << A[i][j];
  }
  cout << endl;
}
}

int main()
{
int n = 0;
cin >> n;

for (int i=0; i<n; ++i)
{
  init();
  input();
  sudoku(0);
  output();
}
return 0;
}