前进方格c++

在一个n*m的方阵中，m为奇数，放置n*m个数，方格中间的下方有一人
用蓝色箭头表示，此人可按照五个方向前进，分别是前面，和前面的相隔的四个，
人每走过一个方格必须取此方格中的数，要求找一条从底到顶的路径，使其数相加之和为最大。输出最大和的值

#include
int main(void)
{
    int array1[6][7] = {
            {16,4,3,12,6,0,3},
            {4,-5,6,7,0,0,2},
            {6,0,-1,-2,3,6,8},
            {5,3,4,0,0,-2,7},
            {-1,7,4,0,7,-5,6},
            {0,-1,3,4,12,4,2}
    };
    int b[6][7], c[6][7];
    int i, j, k;
    int max;
    int flag;

    for (i = 0; i < 6; i++)
        for (j = 0; j < 7; j++)
        {
            b[i][j] = array1[i][j];
            c[i][j] = -1;
        }
    for (i = 1; i < 5; i++)
    {
        for (j = 0; j < 7; j++)
        {
            max = 0;
            for (k = j - 2; k <= j + 2; k++)
            {
                if (k < 0) continue;
                else
                    if (k > 6) break;
                    else
                    {
                        if (b[i][j] + b[i - 1][k] > max)
                        {
                            max = b[i][j] + b[i - 1][k];
                            flag = k;
                        }
                    }
            }
            b[i][j] = max;
            c[i][j] = flag;
        }
    }

    for (j = 1; j <= 5; j++)//i=5
    {
        max = 0;
        for (k = j - 2; k <= j + 2; k++)
        {
            if (k < 0)
                continue;
            else
                if (k > 6)  break;
                else
                {
                    if (b[i][j] + b[i - 1][k] > max)
                    {
                        max = b[i][j] + b[i - 1][k];
                        flag = k;
                    }
                }
        }
        b[i][j] = max;
        c[i][j] = flag;
    }

    max = 0;
    for (j = 1; j <= 5; j++)   //找出
    {
        if (b[i][j] > max)
        {
            max = b[i][j];
            flag = j;
        }
    }
    printf("从底到顶最大和值为：%dnn", max);
    printf("从底到顶分别取数：");
    printf("%d", array1[i][flag]);
    for (j = i; j > 0; j--)
    {
        flag = c[j][flag];
        printf("%5d", array1[j - 1][flag]);
    }
    printf("n");
    return 0;
}