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马踏棋盘 (对应于小甲鱼视频的代码)

马踏棋盘 (对应于小甲鱼视频的代码)

作者: 行走小样 | 来源:发表于2019-02-28 21:08 被阅读0次

    #include<stdio.h>

    #include<time.h>

    #define X 8

    #define Y 8

    int chess[X][Y];

    int c = 1;

    //找到基于(x,y)位置的下一个可走位置

    int nextxy(int *x, int *y, int count)

    {

    switch (count)

    {

    case 0:

    if (*x + 2 <= X - 1 && *y - 1 >= 0 && chess[*x + 2][*y - 1] == 0)

    {

    *x += 2;

    *y -= 1;

    return 1;

    }

    break;

    case 1:

    if (*x + 2 <= X - 1 && *y + 1 <= Y - 1 && chess[*x + 2][*y + 1] == 0)

    {

    *x += 2;

    *y += 1;

    return 1;

    }

    break;

    case 2:

    if (*x - 2 >= 0 && *y - 1 >= 0 && chess[*x - 2][*y - 1] == 0)

    {

    *x -= 2;

    *y -= 1;

    return 1;

    }

    break;

    case 3:

    if (*x - 2 >= 0 && *y + 1 <= Y - 1 && chess[*x - 2][*y + 1] == 0)

    {

    *x -= 2;

    *y += 1;

    return 1;

    }

    break;

    case 4:

    if (*x + 1 <= X - 1 && *y - 2 >= 0 && chess[*x + 1][*y - 2] == 0)

    {

    *x += 1;

    *y -= 2;

    return 1;

    }

    break;

    case 5:

    if (*x + 1 <= X - 1 && *y + 2 <= Y - 1 && chess[*x + 1][*y + 2] == 0)

    {

    *x += 1;

    *y += 2;

    return 1;

    }

    break;

    case 6:

    if (*x - 1 >= 0 && *y - 2 >= 0 && chess[*x - 1][*y - 2] == 0)

    {

    *x -= 1;

    *y -= 2;

    return 1;

    }

    break;

    case 7:

    if (*x - 1 >= 0 && *y + 2 <= Y - 1 && chess[*x - 1][*y + 2] == 0)

    {

    *x -= 1;

    *y += 2;

    return 1;

    }

    break;

    default:

    break;

    }

    return 0;

    }

    void print()

    {

    int i, j;

    for (i = 0; i < X; i++)

    {

    for (j = 0; j < Y; j++)

    {

    printf("%2d\t", chess[i][j]);

    }

    printf("\n");

    }

    printf("\n");

    }

    //深度优先遍历棋盘 (x,y)为起始位置坐标,tag为标记变量,每走一步,tag加一

    int Travalchesboard(int x, int y, int tag)

    {

    int x1 = x, y1 = y, flag = 0, count = 0;

    chess[x][y] = tag;

    if (tag == X * Y)

    {

    //打印棋盘

    print();

    return 1;

    }

    flag = nextxy(&x1, &y1, count);              //打印马的下一个坐标(x1,y1),如果找到flag=1,否则=0;

    while (flag == 0 && count < 7)

    {

    count++;

    flag = nextxy(&x1, &y1, count);

    }

    while (flag)

    {

    if (Travalchesboard(x1, y1, tag + 1))    //这里这个递推需要好好理解一下,在不满足条件出去一个函数后,tag会减一,从而实现不对则往回返

    {

    return 1;

    }

    x1 = x;

    y1 = y;

    count++;

    flag = nextxy(&x1, &y1, count); //出现意外,继续找下一步可走坐标

    }

    if (flag == 0)

    {

    chess[x][y] = 0;

    }

    return 0;

    }

    int main()

    {

    int i, j;

    clock_t start, finish;

    start = clock();

    for (i = 0; i < X; i++)

    {

    for (j = 0; j < Y; j++)

    {

    chess[i][j] = 0;

    }

    }

    if (!Travalchesboard(2, 0, 1))

    {

    printf("sorry,fault!! \n");

    }

    finish = clock();

    printf("\n time: \n", (double)(finish - start) / CLOCKS_PER_SEC);

    return 0;

    }

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