我花了几天时间尝试制作一个数独解决程序，但我被这些方法所困扰。我在这里找到了这个算法，但我不太理解它：
- 从第一个空单元格开始，并在其中输入 1。
- 检查整个板子，看看是否有冲突
- 如果板上存在冲突，请将当前单元格中的数字加 1（因此将 1 更改为 2、2 更改为 3 等）
- 如果棋盘是干净的，则再次从第一步开始。
- 如果给定单元格上的所有九个可能的数字都会导致棋盘中发生冲突，则您可以将此单元格设置回空，返回到前一个单元格，然后从步骤 3 重新开始（这就是“回溯”的用武之地）。
这是我的代码。我觉得我的有问题帮助解决（...）功能。您能帮我找出问题所在吗？
#include <iostream>
#include <iomanip>
#include <time.h>
#include <cstdlib>
#include <windows.h>
using namespace std;
class Sudoku
{
private:
int board[9][9];
int change[9][9];
public:
Sudoku();
void Print_Board();
void Add_First_Cord();
void Solve();
void Help_Solve(int i, int j);
bool Check_Conflicts(int p, int i, int j);
};
Sudoku Game;
void setcolor(unsigned short color) //The function that you'll use to
{ //set the colour
HANDLE hcon = GetStdHandle(STD_OUTPUT_HANDLE);
SetConsoleTextAttribute(hcon,color);
}
Sudoku::Sudoku()
{
for(int i = 1; i <= 9; i++)
for(int j = 1; j <= 9; j++)
board[i][j] = 0;
}
void Sudoku::Print_Board()
{
for(int i = 1; i <= 9; i++)
{
for(int j = 1; j <= 9; j++)
{
if(change[i][j] == 1)
{
setcolor(12);
cout << board[i][j] << " ";
setcolor(7);
}
else cout << board[i][j] << " ";
if(j%3 == 0) cout << "| ";
}
cout << endl;
if(i%3 == 0) cout << "------+-------+---------" << endl;
}
}
void Sudoku::Add_First_Cord()
{
board[1][1] = 5; change[1][1] = 1;
board[1][2] = 3; change[1][2] = 1;
board[1][5] = 7; change[1][5] = 1;
board[2][1] = 6; change[2][1] = 1;
board[2][4] = 1; change[2][4] = 1;
board[2][5] = 9; change[2][5] = 1;
board[2][6] = 5; change[2][6] = 1;
board[3][2] = 9; change[3][2] = 1;
board[3][3] = 8; change[3][3] = 1;
board[3][8] = 6; change[3][8] = 1;
board[4][1] = 8; change[4][1] = 1;
board[4][5] = 6; change[4][5] = 1;
board[4][9] = 3; change[4][9] = 1;
board[5][1] = 4; change[5][1] = 1;
board[5][4] = 8; change[5][4] = 1;
board[5][6] = 3; change[5][6] = 1;
board[5][9] = 1; change[5][9] = 1;
board[6][1] = 7; change[6][1] = 1;
board[6][5] = 2; change[6][5] = 1;
board[6][9] = 6; change[6][9] = 1;
board[7][2] = 6; change[7][2] = 1;
board[7][7] = 2; change[7][7] = 1;
board[7][8] = 8; change[7][8] = 1;
board[8][4] = 4; change[8][4] = 1;
board[8][5] = 1; change[8][5] = 1;
board[8][6] = 9; change[8][6] = 1;
board[8][9] = 5; change[8][9] = 1;
board[9][5] = 8; change[9][5] = 1;
board[9][8] = 7; change[9][8] = 1;
board[9][9] = 9; change[9][9] = 1;
}
bool Sudoku::Check_Conflicts(int p, int i, int j)
{
for(int k = 1; k <= 9; k++)
if(board[i][k] == p) return false;
for(int q = 1; q <= 9; q++)
if(board[q][j] == p) return false;
/*
*00
000
000
*/
if((j == 1 || j == 4 || j == 7) && (i == 1 || i == 4 || i == 7))
{
if(board[i][j+1] == p || board[i][j+2] == p || board[i+1][j] == p ||
board[i+2][j] == p || board[i+1][j+1] == p || board[i+1][j+2] == p ||
board[i+2][j+1] == p || board[i+2][j+2] == p)return false;
}
/*
000
000
*00
*/
if((j == 1 || j == 4 || j == 7) && (i == 3 || i == 6 || i == 9))
{
if(board[i-1][j] == p || board[i-2][j] == p || board[i][j+1] == p ||
board[i][j+2] == p || board[i-1][j+1] == p || board[i-1][j+2] == p ||
board[i-2][j+1] == p || board[i-2][j+2] == p)return false;
}
/*
000
*00
000
*/
if((j == 1 || j == 4 || j == 7) && (i == 2 || i == 5 || i == 8))
{
if(board[i-1][j] == p || board[i+1][j] == p || board[i-1][j+1] == p ||
board[i][j+1] == p || board[i+1][j+1] == p || board[i+1][j+2] == p ||
board[i][j+2] == p || board[i+1][j+2] == p)return false;
}
/*
0*0
000
000
*/
if((j == 2 || j == 5 || j == 8) && (i == 1 || i == 5 || i == 7))
{
if(board[i-1][j] == p || board[i+1][j] == p || board[i-1][j+1] == p ||
board[i][j+1] == p || board[i+1][j+1] == p || board[i+1][j+2] == p ||
board[i][j+2] == p || board[i+1][j+2] == p)return false;
}
/*
000
0*0
000
*/
if((j == 2 || j == 5 || j == 8) && (i == 2 || i == 5 || i == 8))
{
if(board[i-1][j] == p || board[i-1][j-1] == p || board[i-1][j+1] == p ||
board[i][j+1] == p || board[i][j-1] == p || board[i+1][j+1] == p ||
board[i][j] == p || board[i+1][j-1] == p)return false;
}
/*
000
000
0*0
*/
if((j == 2 || j == 5 || j == 8) && (i == 3 || i == 6 || i == 9))
{
if(board[i][j-1] == p || board[i][j+1] == p || board[i-1][j] == p ||
board[i-1][j+1] == p || board[i-1][j-1] == p || board[i-2][j] == p ||
board[i-1][j+1] == p || board[i-2][j-1] == p) return false;
}
/*
00*
000
000
*/
if((j == 3 || j == 6 || j == 9) && (i == 1 || i == 4 || i == 7))
{
if(board[i][j-1] == p || board[i][j-2] == p || board[i+1][j] == p ||
board[i+1][j-1] == p || board[i+1][j-2] == p || board[i+2][j] == p ||
board[i+2][j-1] == p || board[i+2][j-2] == p) return false;
}
/*
000
00*
000
*/
if((j == 3 || j == 6 || j == 9) && (i == 2 || i == 5 || i == 8))
{
if(board[i-1][j] == p || board[i-1][j-1] == p || board[i-1][j-2] == p ||
board[i][j-1] == p || board[i][j-2] == p || board[i+1][j] == p ||
board[i+1][j-1] == p || board[i+1][j-2] == p) return false;
}
/*
000
000
00*
*/
if((j == 3 || j == 6 || j == 9) && (i == 3 || i == 6 || i == 9))
{
if(board[i][j-1] == p || board[i][j-1] == p || board[i-1][j] == p ||
board[i-1][j-1] == p || board[i-1][j-2] == p || board[i-2][j] == p ||
board[i-2][j-1] == p || board[i-2][j-2] == p) return false;
}
return true;
}
void Sudoku::Help_Solve(int i, int j)
{
if(j <= 0)
{
i = i-1;
j = 9;
}
if(change[i][j] == 1) return Game.Help_Solve(i, j-1);
for(int p = 1; p <= 9; p++)
if(Game.Check_Conflicts(p, i, j))
{
board[i][j] = p;
return;
}
return Game.Help_Solve(i, j-1);
}
void Sudoku::Solve()
{
for(int i = 1; i <= 9; i++)
{
for(int j = 1; j <= 9; j++)
{
if(board[i][j] == 0 && change[i][j] == 0)
{
Game.Help_Solve(i, j);
}
}
}
for(int i = 1; i <= 9; i++)
for(int j = 1; j <= 9; j++)
if(board[i][j] == 0) Game.Help_Solve(i, j);
}
int main()
{
Game.Add_First_Cord();
Game.Solve();
Game.Print_Board();
system("pause");
return 0;
}
Edit: I need to use recursion right? But maybe the parameters I give to the function are wrong. I really don't know. In Add_First_Cord() I declare the starting values that every sudoku has in the beginning. Here are the values that I use: http://bg.wikipedia.org/wiki/%D0%A4%D0%B0%D0%B9%D0%BB:Sudoku-by-L2G-20050714.gif http://bg.wikipedia.org/wiki/%D0%A4%D0%B0%D0%B9%D0%BB:Sudoku-by-L2G-20050714.gif. I expect to see the solved sudoku as it is shown in wikipedia. But some solved values are right others are not. Here is what I get in the console 
TNode
TNode => vector<TNode>
S can have number N added at row I, column J if:
(I,J)是空的N in row I
N在列中J
N在包含 3x3 的正方形中(I,J)
S to the vector满足这些规则的州您当前的方法混合了要搜索的图的规范和搜索算法的实现。如果你把这两者混合起来，你会遇到很多困难。这个问题自然地分为两个不同的部分——算法和图表——所以你可以而且应该在你的实现中利用它。这将使事情变得更加简单。
如果你选择这种分离，你得到的另一个好处是你将能够reuse你的图搜索算法解决了大量问题 - 非常酷！