c - 矩阵中带有障碍物的路径数

Question

所以问题是关于打印矩阵从左上角到右下角的所有路径。 允许的移动是向下、向右和对角线。 这必须应用于障碍物，即如果存在障碍物，则不应打印该路径。所以我写了这段代码，如果运行输出是

1 4 7 8 9

1 4 5 8 9

1 4 5 8 9

1 4 5 9

1 4 8 9

1 5 8 9

1 5 8 9

1 5 9

第 3 行和第 7 行的最后一秒值需要为 6。但相反，它是 8。有什么想法吗？蒂亚

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

void paths(int *mat, int i, int j, int n, int *path, int val)
{
    int k,l;
    if (i == n - 1)
    {
        for(k = j; k < n; k++)
        {path[val + k - j] = *((mat + i*n) + k);}
        for(l = 0; l < val + n - j; l++)
        {if(path[l] == 0){return;}}
        for(l = 0; l < val + n - j; l++)
        {
            printf("%d ", path[l]);
        }printf("n");
        return;
    }
    if (j == n - 1)
    {
        for(k = j; k < n; k++)
        {path[val + k - i] = *((mat + k*n) + j);}
        for(l = 0; l < val + n - j; l++)
        {if(path[l] == 0){return;}}
        for(l = 0; l < val + n - i; l++)
        {
            printf("%d ", path[l]);
        }printf("n");
        return;
    }
    path[val] = *((mat + i*n) + j);
    paths(mat, i+1, j, n, path, val+1);
    paths(mat, i, j+1, n, path, val+1);
    paths(mat, i+1, j+1, n, path, val+1);
}
void printAll(int *mat, int n)
{
    int *path = malloc((2*n) * sizeof(int));
    paths(mat, 0, 0, n, path, 0);
}
int main()
{
    int n,i,j; 
    printf("Enter the size of the gridn");
    scanf("%d", &n);
    int mat[n][n];
    int count = 1;
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < n; j++)
        {
            mat[i][j] = count; count = count + 1;
        }
    }
    printf("Enter the grid points that are offsetsn");
    int obs[n*n][2];i=0;
    while(1 == 1)
    {
        scanf("%d%d", &obs[i][0], &obs[i][1]);
        mat[obs[i][0]][obs[i][1]] = 0;
        if(obs[i][0] == -1 || obs[i][1] == -1){break;}
        i = i + 1;
    }
    printf("The paths for the robot aren");
    printAll(*mat, n);
}

Answer 1

要获得正确的输出，请将 paths() 函数的 for 循环更改为：

for(k = i; k < n; k++)
        {path[val + k - j] = *((mat + i*n) + k);}