数独

数独求解

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#include <stdio.h>
#define UNASSIGNED 0
#define N 9 // NxN 的大小
// 是否还有未放置的位置
int FindUnassignedLocation(int grid[N][N], int *row, int *col);
// 在位置 grid[row][col]处,是否可以放置num
int isSafe(int grid[N][N], int row, int col, int num);
int SolveSudoku(int grid[N][N]) // 回溯法求解
{
int row, col;
int *p1,*p2;
p1=&row;
p2=&col;
if (FindUnassignedLocation(grid, p1, p2)==0)//都已经放置完,则返true
return 1; // success!
for (int num = 1; num <= 9; num++)
{
if (isSafe(grid, row, col, num))
{
grid[row][col] = num; // 放置棋子
if (SolveSudoku(grid)) // return, if success, yay!
return 1;
grid[row][col] = UNASSIGNED; // 失败, 设置为 未放置的 & 再一次尝试
}
}
return 0;
}
int FindUnassignedLocation(int grid[N][N], int *row, int *col)
{
for (*row = 0; *row < N; (*row)++)
for (*col = 0; *col < N; (*col)++)
if (grid[*row][*col] == UNASSIGNED)
return 1;
return 0;
}
int UsedInRow(int grid[N][N], int row, int num)
{
for (int col = 0; col < N; col++)
if (grid[row][col] == num)
return 1;
return 0;
}
int UsedInCol(int grid[N][N], int col, int num)
{
for (int row = 0; row < N; row++)
if (grid[row][col] == num)
return 1;
return 0;
}
int UsedInBox(int grid[N][N], int boxStartRow, int boxStartCol, int num)
{
for (int row = 0; row < 3; row++)
for (int col = 0; col < 3; col++)
if (grid[row+boxStartRow][col+boxStartCol] == num)
return 1;
return 0;
}
int isSafe(int grid[N][N], int row, int col, int num)
{
return UsedInRow(grid, row, num)==0 &&UsedInCol(grid, col, num)==0 &&UsedInBox(grid, row - row%3 , col - col%3, num)==0;
}
void printGrid(int grid[N][N])
{
for (int row = 0; row < N; row++)
{
for (int col = 0; col < N; col++)
printf("%4d", grid[row][col]);
printf("\n\n");
}
}
/* 测试 */
int main()
{
int grid[N][N] = {{3, 0, 6, 5, 0, 8, 4, 0, 0},
{5, 2, 0, 0, 0, 0, 0, 0, 0},
{0, 8, 7, 0, 0, 0, 0, 3, 1},
{0, 0, 3, 0, 1, 0, 0, 8, 0},
{9, 0, 0, 8, 6, 3, 0, 0, 5},
{0, 5, 0, 0, 9, 0, 6, 0, 0},
{1, 3, 0, 0, 0, 0, 2, 5, 0},
{0, 0, 0, 0, 0, 0, 0, 7, 4},
{0, 0, 5, 2, 0, 6, 3, 0, 0}};
if (SolveSudoku(grid))
printGrid(grid);
else
printf("No solution exists");
return 0;
}

运行结果截图:
"数独求解"