数独算法c语言,数独求解算法C.
我正在努力制作一个Sudoku Solving程序几天,但我坚持使用这些方法.我在这里找到了这个算法,但我真的不明白它:
start at the first empty cell, and put 1 in it.
Check the entire board, and see if there are any conflicts
If there are coflicts on the board, increase the number in the current cell by 1 (so change 1 to 2, 2 to 3, etc)
If the board is clean move, start at step one again.
If all nine possible numbers on a given cell cause a conflict in the board, then you set this cell back to empty, go back to the previous cell, and start again from step 3 (this is where the ‘backtracking’ comes in).
这是我的代码.我认为我的Help_Solve(…)函数出了问题.你能帮我解决一下这个问题吗?
#include
#include
#include
#include
#include
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;
}
编辑:我需要使用递归吗?但也许我给函数的参数是错误的.我真的不知道.在Add_First_Cord()中,我声明了每个数独在开头具有的起始值.以下是我使用的值:http://bg.wikipedia.org/wiki/%D0%A4%D0%B0%D0%B9%D0%BB:Sudoku-by-L2G-20050714.gif.我希望看到解析的数独,因为它在维基百科中显示.但有些解决的价值是正确的,其他则不是.这是我在控制台中得到的
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