简体   繁体   English

C编程——数独解决方案(回溯)

[英]C programming - Sodoku solution (backtracking)

the Question is: to get from the user sodoku board and if there is a solution to print it, if not to print no solution: the solution of the soduko;问题是:从用户sodoku board获得,如果有解决方案打印它,如果没有打印没有解决方案:soduko的解决方案; two identical numbers mmust not appear on the same line.两个相同的数字不得出现在同一行。 two identical numbers must not appear in the same colum.两个相同的数字不能出现在同一列中。 I worte a program that works perfectly when I put the soduko board and the size (global parametes-as shown un my code) but when I tried to receive from the user it took so much time to run the solution and sometimes it didn't retun anything?我编写了一个程序,当我放置 soduko 板和尺寸(全局参数 - 如我的代码所示)时,我编写了一个运行良好的程序,但是当我尝试从用户那里接收时,运行解决方案花费了很多时间,有时它没有退还什么? I would like to understand why?!我想知道为什么?!

#include <stdio.h>
#include <stdlib.h>

#define SIZE 5


int matrix[5][5] = {
    {4,2,0,0,5},
    {2,0,0,1,3},
    {5,0,1,2,0},
    {0,0,3,0,2},
    {0,0,0,0,0},
};


void print_sudoku()
{
    int i,j;
    for(i=0;i<SIZE;i++)
    {
        for(j=0;j<SIZE;j++)
        {
            printf("%d ",matrix[i][j]);
        }
        printf("\n");
    }
}


int number_unassigned(int *row, int *col)
{
    int num_unassign = 0;
    int i,j;
    for(i=0;i<SIZE;i++)
    {
        for(j=0;j<SIZE;j++)
        {
            if(matrix[i][j] == 0)
            {
                *row = i;
                *col = j;
                num_unassign = 1;
                return num_unassign;
            }
        }
    }
    return num_unassign;
}


int is_safe(int n, int r, int c)
{
    int i;

    for(i=0;i<SIZE;i++)
    {
        if(matrix[r][i] == n)
            return 0;
    }
    for(i=0;i<SIZE;i++)
    {
        if(matrix[i][c] == n)
            return 0;
    }
    return 1;
}


int solve_sudoku()
{
    int row;
    int col;
 
    if(number_unassigned(&row, &col) == 0)
        return 1;

    int i;

    for(i=1;i<=SIZE;i++)
    {
  
        if(is_safe(i, row, col))
        {
            matrix[row][col] = i;

            if(solve_sudoku())
                return 1;

            matrix[row][col]=0;
        }
    }
    return 0;
}


int main()
{
    if (solve_sudoku())
        print_sudoku();
    else
        printf("No solution!\n");
    return 0;
}

and this is whe code that I used to ask the user to enter a sodoku board:这就是我用来要求用户输入数独板的代码:

int** ReadSoduko(int n) {
    int** matrix = (int**) malloc((sizeof(int*)) * n);
    for(int i = 0; i < n; ++i) {
        matrix[i] = malloc(sizeof(int) * n);
    }
    printf("\nEnter your soduko board:\n");
    for(int i = 0; i < n; ++i) {
          printf("Enter row [%d]: ", i);
        for(int j = 0; j < n; ++j) {
            scanf("%d", &matrix[i][j]);
        }
    }
    return matrix;
}

I suspect your problem is related to one thing: #define SIZE that you are probably forgetting to update when reading dynamically.我怀疑您的问题与一件事有关: #define SIZE您可能在动态阅读时忘记更新。 Since you use SIZE in your loops, if that is not the real size of your matrix, then it probably won't work.由于您在循环中使用了SIZE ,如果这不是矩阵的实际大小,那么它可能不起作用。 I've changed 2 lines and added 3 other (the lines with comments at the end).我更改了 2 行并添加了其他 3 行(末尾带有注释的行)。 Try it now.现在就试试。

#include <stdio.h>
#include <stdlib.h>

int SIZE; //changed here

int** matrix; //changed here

void print_sudoku()
{
    int i, j;
    for (i = 0; i < SIZE; i++)
    {
        for (j = 0; j < SIZE; j++)
        {
            printf("%d ", matrix[i][j]);
        }
        printf("\n");
    }
}


int number_unassigned(int* row, int* col)
{
    int num_unassign = 0;
    int i, j;
    for (i = 0; i < SIZE; i++)
    {
        for (j = 0; j < SIZE; j++)
        {
            if (matrix[i][j] == 0)
            {
                *row = i;
                *col = j;
                num_unassign = 1;
                return num_unassign;
            }
        }
    }
    return num_unassign;
}


int is_safe(int n, int r, int c)
{
    int i;

    for (i = 0; i < SIZE; i++)
    {
        if (matrix[r][i] == n)
            return 0;
    }
    for (i = 0; i < SIZE; i++)
    {
        if (matrix[i][c] == n)
            return 0;
    }
    return 1;
}


int solve_sudoku()
{
    int row;
    int col;

    if (number_unassigned(&row, &col) == 0)
        return 1;

    int i;

    for (i = 1; i <= SIZE; i++)
    {

        if (is_safe(i, row, col))
        {
            matrix[row][col] = i;

            if (solve_sudoku())
                return 1;

            matrix[row][col] = 0;
        }
    }
    return 0;
}


int** ReadSoduko(int n) {
    int** matrix = (int**)malloc((sizeof(int*)) * n);
    for (int i = 0; i < n; ++i) {
        matrix[i] = (int*) malloc(sizeof(int) * n);
    }
    printf("\nEnter your soduko board:\n");
    for (int i = 0; i < n; ++i) {
        printf("Enter row [%d]: ", i);
        for (int j = 0; j < n; ++j) {
            scanf("%d", &matrix[i][j]);
        }
    }
    return matrix;
}

int main()
{
    printf("Size of matrix: "); //added this
    scanf("%d", &SIZE); //added this
    matrix = ReadSoduko(SIZE); //added this
    if (solve_sudoku())
        print_sudoku();
    else
        printf("No solution!\n");
    return 0;
}

And don't forget that the matrix that you have declared statically doesn't have a solution, I have tested the same matrix, just replacing the 2 in first line by a 0 and it worked here.并且不要忘记您静态声明的矩阵没有解决方案,我已经测试了相同的矩阵,只是replacing the 2 in first line by a 0并且它在这里工作。

声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM