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Linux环境下连连看游戏代码,《连连看》算法C语言演示(自动连连看)

基于gcc的连连看小游戏

看题目就知道是写给初学者的,没需要的就别看了,自己都觉得怪无聊的。

很多游戏的耐玩性都来自精巧的算法,特别是人工智能的水平。比如前几天看了著名的Alpha GO的算法,用了复杂的人工智能网络。而最简单的,可能就是连连看了,所以很多老师留作业,直接就是实现连连看。

连连看游戏的规则非常简单:

两个图片相同。

两个图片之间,沿着相邻的格子画线,中间不能有障碍物。

画线中间最多允许2个转折。

所以算法主要是这样几部分:

用数据结构描述图板。很简单,一个2维的整数数组,数组的值就是图片的标志,相同的数字表示相同的图片。有一个小的重点就是,有些连连看的地图中,允许在边界的两个图片,从地图外连线消除。这种情况一般需要建立的图板尺寸,比实际显示的图板,周边大一个格子,从而描述可以连线的空白外边界。本例中只是简单的使用完整的图板,不允许利用边界外连线。

生成图板。通常用随机数产生图片ID来填充图板就好。比较复杂的游戏,会有多种的布局方式,例如两个三角形。这种一般要手工编辑图板模板,在允许填充的区域事先用某个特定的整数值来标注,随后的随机数填充只填充允许填充的区域。本例中只是简单的随机填充。

检查连线中的障碍物。确定有障碍物的关键在于确定什么样的格子是空。通常定义格子的值为0就算空。要求所有的图片ID从1开始顺序编码。复杂的游戏还会定义负数作为特定的标志,比如允许填充区之类的。

检查直接连接:两张图片的坐标,必然x轴或者y轴有一项相同,表示两张图片在x轴或者y轴的同一条线上才可能出现直接连接。随后循环检查两者之间是否有障碍物即可确定。

检查一折连接:与检查直接连接相反,两个图片必须不在一条直线上,才可能出现一折连接,也就是x/y必须都不相同。随后以两张图片坐标,可以形成一个矩阵,矩阵的一对对角是两张图片,假设是A/B两点。矩阵另外两个对角分别是C1/C2,分别检查A/C1和C1/B或者A/C2和C2/B能同时形成直线连接,则A图片到B图片的1折连接可以成立。描述比较苍白,建议你自己画张简单的图就容易理解了。在一折连接的检查中,会调用上面的直线连接的检测至少2次,这种调用的方式有点类似递归的调用。

检查两折连接:同样假设两张图片分别为A/B两点,在A点的X+/X-方向/Y+方向/Y-方向,共4个方向上循环查找是否存在一个点C,使得A到C为直线连接,C到B为1折连接,则两折连接成立。这中间,会调用前面的直接连接检测和一折连接检测。

用到的算法基本就是这些,下面看程序。本程序使用GCC或者CLANG编译的,可以在Linux或者Mac直接编译执行。

#include

#include

#include

#include

//常量习惯定义在程序一开始,以便将来的修改,比如重新定义一个更大的地图界限

//定义图板尺寸

#define _width 20

#define _height 20

//定义数组矩阵中,0表示该格子为空

#define empty (0)

//定义共有20种图片

#define _pics (20)

//定义在图板中随机产生100*2个图片的填充

//使用100是为了每次产生2个相同的图片,从而保证整个图可以消除完

#define _datas (100)

//C语言没有bool类型,为了方便自定义一个

typedef int bool;

#define TRUE (1)

#define FALSE (0)

//定义一个结构用来描述一个点坐标

typedef struct {

int x;

int y;

} _point;

//描述图板的数组

int map[_width][_height];

//-------------------------init map----------------------

//从图板中获取一个空白格子的坐标,这种方法随着填充图片的增加,

//效率会急剧降低,不过简单实用,这么小的图板对cpu来说也不算什么

_point getRndEmptyBox(){

int x,y;

while(TRUE){

//gcc的随机数跟windows的随机数产生规则不同

//linux是产生从0开始到RAND_MAX的一个正整数

//如果移植到windows,这部分要修改

int x=rand() % _width;

int y=rand() % _height;

if (map[x][y]==empty){

_point r;

r.x=x;

r.y=y;

return r;

}

}

}

//设置一对随机图片

void setRandPic(){

_point p;

//+1是为了防止出现随机数为0的情况,那样等于填充了空白

int pic=rand() % _pics + 1;

p = getRndEmptyBox();

map[p.x][p.y]=pic;

//printf("[%02d,%02d]=%02d\n",p.x,p.y,pic);

p = getRndEmptyBox();

map[p.x][p.y]=pic;

return;

}

//用随机图片填充整个图板

void setRndMap(){

int i;

for(i=0;i<_datas>

setRandPic();

}

return;

}

//-----------------------------show status --------------------

//显示当前的图板情况

void dumpMap(){

int i,j;

printf("--: ");

for(i=0;i<_width>

printf("%02d ",i);

}

printf("\n");

for(i=0;i<_height>

printf("%02d: ",i);

for(j=0;j<_width>

printf("%02d ",map[j][i]);

}

printf("\n");

}

}

//显示当前的图板情况,并且使用红色标注上将要消除的2个点

//显示部分使用了linux的终端控制专用方式,移植到windows时需要修改

void dumpMapWithHotPoint(_point c1,_point c2){

int x,y;

//为了方便计数,显示x/y轴格子编号

printf("--: ");

for(x=0;x<_width>

printf("%02d ",x);

}

printf("\n");

for(y=0;y<_height>

printf("%02d: ",y);

for(x=0;x<_width>

if ((c1.x==x && c1.y==y) || (c2.x==x && c2.y==y))

printf("\e[1;31m%02d\e[0m ",map[x][y]);

else

printf("%02d ",map[x][y]);

}

printf("\n");

}

}

//-------------------------search path--------------------

//检查直接连接,返回成功或者失败

bool havePathCorner0(_point p1,_point p2){

if (p1.x != p2.x && p1.y != p2.y)

return FALSE; // not in the same line

int min,max;

if (p1.x == p2.x){

min = p1.y < p2.y ? p1.y : p2.y;

max = p1.y > p2.y ? p1.y : p2.y;

for(min++;min < max;min++){

if(map[p1.x][min] != empty)

return FALSE;  //have block false

}

} else {

min = p1.x < p2.x ? p1.x : p2.x;

max = p1.x > p2.x ? p1.x : p2.x;

for(min++;min < max;min++){

if(map[min][p1.y] != empty)

return FALSE; //have block false

}

}

return TRUE;

}

//检查1折连接,返回1个点,

//如果点的坐标为负表示不存在1折连接

_point havePathCorner1(_point p1,_point p2){

_point nullPoint;

nullPoint.x=nullPoint.y=-1;

if (p1.x == p2.x || p1.y == p2.y)

return nullPoint;

_point c1,c2;

c1.x=p1.x;

c1.y=p2.y;

c2.x=p2.x;

c2.y=p1.y;

if (map[c1.x][c1.y] ==  empty){

bool b1=havePathCorner0(p1,c1);

bool b2=havePathCorner0(c1,p2);

if (b1 && b2)

return c1;

}

if (map[c2.x][c2.y] ==  empty){

bool b1=havePathCorner0(p1,c2);

bool b2=havePathCorner0(c2,p2);

if (b1 && b2)

return c2;

}

return nullPoint;

}

//检查两折连接,返回两个点,

//返回点坐标为负表示不存在两折连接

//其中使用了4个方向的循环查找

_point result[2];

_point *havePathCorner2(_point p1,_point p2){

int i;

_point *r=result;

//search direction 1

for(i=p1.y+1;i<_height>

if (map[p1.x][i] == empty){

_point c1;

c1.x=p1.x;

c1.y=i;

_point d1=havePathCorner1(c1,p2);

if (d1.x != -1){

r[0].x=c1.x;

r[0].y=c1.y;

r[1].x=d1.x;

r[1].y=d1.y;

return r;

}

} else

break;

}

//search direction 2

for(i=p1.y-1;i>-1;i--){

if (map[p1.x][i] == empty){

_point c1;

c1.x=p1.x;

c1.y=i;

_point d1=havePathCorner1(c1,p2);

if (d1.x != -1){

r[0].x=c1.x;

r[0].y=c1.y;

r[1].x=d1.x;

r[1].y=d1.y;

return r;

}

} else

break;

}

//search direction 3

for(i=p1.x+1;i<_width>

if (map[i][p1.y] == empty){

_point c1;

c1.x=i;

c1.y=p1.y;

_point d1=havePathCorner1(c1,p2);

if (d1.x != -1){

r[0].x=c1.x;

r[0].y=c1.y;

r[1].x=d1.x;

r[1].y=d1.y;

return r;

}

} else

break;

}

//search direction 4

for(i=p1.x-1;i>-1;i--){

if (map[i][p1.y] == empty){

_point c1;

c1.x=i;

c1.y=p1.y;

_point d1=havePathCorner1(c1,p2);

if (d1.x != -1){

r[0].x=c1.x;

r[0].y=c1.y;

r[1].x=d1.x;

r[1].y=d1.y;

return r;

}

} else

break;

}

r[1].x=r[0].x=r[0].y=r[1].y=-1;

return r;

}

//汇总上面的3种情况,查找两个点之间是否存在合法连接

bool havePath(_point p1,_point p2){

if (havePathCorner0(p1,p2)){

printf("[%d,%d] to [%d,%d] have a direct path.\n",p1.x,p1.y,p2.x,p2.y);

return TRUE;

}

_point r=havePathCorner1(p1,p2);

if (r.x != -1){

printf("[%d,%d] to [%d,%d] have a 1 cornor path throught [%d,%d].\n",

p1.x,p1.y,p2.x,p2.y,r.x,r.y);

return TRUE;

}

_point *c=havePathCorner2(p1,p2);

if (c[0].x != -1){

printf("[%d,%d] to [%d,%d] have a 2 cornor path throught [%d,%d] and [%d,%d].\n",

p1.x,p1.y,p2.x,p2.y,c[0].x,c[0].y,c[1].x,c[1].y);

return TRUE;

}

return FALSE;

}

//对于给定的起始点,查找在整个图板中,起始点之后的所有点,

//是否存在相同图片,并且两张图片之间可以合法连线

bool searchMap(_point p1){

int ix,iy;

bool inner1=TRUE;

//printf("begin match:%d,%d\n",p1.x,p1.y);

int c1=map[p1.x][p1.y];

for (iy=p1.y;iy<_height>

for(ix=0;ix<_width>

//遍历查找整个图板的时候,图板中,起始点之前的图片实际已经查找过

//所以应当从图片之后的部分开始查找才有效率

//遍历的方式是逐行、每行中逐个遍历

//在第一次循环的时候,x坐标应当也是起始点的下一个,所以使用inner1来确认第一行循环

if (inner1){

ix=p1.x+1;

inner1=FALSE;

}

if(map[ix][iy] != c1){

//printf("skip:%d,%d\n",ix,iy);

//continue;

} else {

_point p2;

p2.x=ix;

p2.y=iy;

if (!havePath(p1,p2)){

//printf("No path from [%d,%d] to [%d,%d]\n",p1.x,p1.y,p2.x,p2.y);

} else {

dumpMapWithHotPoint(p1,p2);

map[p1.x][p1.y]=empty;

map[p2.x][p2.y]=empty;

//dumpMap();

return TRUE;

}

}

}

};

return FALSE;

}

//这个函数式扫描全图板,自动连连看

bool searchAllMap(){

int ix,iy;

bool noPathLeft=FALSE;

while(!noPathLeft){

noPathLeft=TRUE;

for (iy=0;iy<_height>

for(ix=0;ix<_width>

if(map[ix][iy] != empty){

_point p;

p.x=ix;

p.y=iy;

if(searchMap(p) && noPathLeft)

noPathLeft=FALSE;

}

}

}

printf("next loop...\n");

};

return TRUE;

}

//-----------------main-----------------------------

int main(int argc,char **argv){

srand((unsigned)time(NULL));

memset(map,0,sizeof(map));

setRndMap();

dumpMap();

searchAllMap();

}

运行结果会是类似这样:

link> ./linktest

--: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19

00: 14 00 02 16 18 06 00 00 00 00 00 12 13 04 00 00 00 19 18 00

01: 00 10 00 12 00 05 00 00 00 00 00 00 00 15 09 00 00 00 18 00

02: 00 00 03 00 00 13 16 00 05 17 00 17 00 00 07 05 00 00 05 16

03: 02 00 00 00 00 13 00 17 15 00 00 00 00 00 00 02 00 11 15 08

04: 05 11 00 08 05 00 06 00 00 00 07 06 00 00 06 00 15 17 00 00

05: 17 18 16 11 01 04 00 16 18 00 04 01 00 02 19 18 00 11 16 00

06: 00 01 00 11 00 00 00 12 03 00 02 17 01 00 00 19 00 13 07 03

07: 06 10 00 10 10 00 00 02 00 00 11 15 09 18 00 00 00 00 07 00

08: 09 14 06 19 00 09 00 00 09 18 00 00 00 12 18 05 00 11 00 18

09: 01 00 00 07 06 00 15 00 00 00 00 00 00 02 11 00 00 00 08 00

10: 00 00 02 03 00 15 00 00 19 00 00 07 00 12 00 00 10 00 19 00

11: 12 11 14 09 10 00 00 00 19 18 00 13 05 11 05 00 00 18 00 07

12: 11 00 09 00 00 00 00 10 03 00 00 00 00 00 00 00 16 05 12 00

13: 02 17 00 05 00 00 00 00 04 00 07 00 01 00 09 00 00 00 19 00

14: 07 00 00 17 00 00 06 00 00 14 00 00 05 00 09 00 08 00 18 00

15: 00 02 19 00 04 16 00 00 14 00 00 15 16 14 00 00 00 00 00 12

16: 00 02 00 16 09 00 00 00 00 00 00 09 13 01 19 15 00 17 00 15

17: 00 18 00 00 08 00 00 00 10 00 00 00 00 06 00 09 02 06 00 01

18: 00 00 15 00 00 02 08 00 09 07 00 18 06 00 09 00 11 00 00 15

19: 06 18 00 00 00 02 17 00 00 00 19 00 19 00 00 00 00 04 00 03

[18,0] to [18,1] have a direct path.

--: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19

00: 14 00 02 16 18 06 00 00 00 00 00 12 13 04 00 00 00 19 18 00

01: 00 10 00 12 00 05 00 00 00 00 00 00 00 15 09 00 00 00 18 00

02: 00 00 03 00 00 13 16 00 05 17 00 17 00 00 07 05 00 00 05 16

03: 02 00 00 00 00 13 00 17 15 00 00 00 00 00 00 02 00 11 15 08

04: 05 11 00 08 05 00 06 00 00 00 07 06 00 00 06 00 15 17 00 00

05: 17 18 16 11 01 04 00 16 18 00 04 01 00 02 19 18 00 11 16 00

06: 00 01 00 11 00 00 00 12 03 00 02 17 01 00 00 19 00 13 07 03

07: 06 10 00 10 10 00 00 02 00 00 11 15 09 18 00 00 00 00 07 00

08: 09 14 06 19 00 09 00 00 09 18 00 00 00 12 18 05 00 11 00 18

09: 01 00 00 07 06 00 15 00 00 00 00 00 00 02 11 00 00 00 08 00

10: 00 00 02 03 00 15 00 00 19 00 00 07 00 12 00 00 10 00 19 00

11: 12 11 14 09 10 00 00 00 19 18 00 13 05 11 05 00 00 18 00 07

12: 11 00 09 00 00 00 00 10 03 00 00 00 00 00 00 00 16 05 12 00

13: 02 17 00 05 00 00 00 00 04 00 07 00 01 00 09 00 00 00 19 00

14: 07 00 00 17 00 00 06 00 00 14 00 00 05 00 09 00 08 00 18 00

15: 00 02 19 00 04 16 00 00 14 00 00 15 16 14 00 00 00 00 00 12

16: 00 02 00 16 09 00 00 00 00 00 00 09 13 01 19 15 00 17 00 15

17: 00 18 00 00 08 00 00 00 10 00 00 00 00 06 00 09 02 06 00 01

18: 00 00 15 00 00 02 08 00 09 07 00 18 06 00 09 00 11 00 00 15

19: 06 18 00 00 00 02 17 00 00 00 19 00 19 00 00 00 00 04 00 03

......

[10,17] to [19,18] have a 1 cornor path throught [10,18].

--: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19

00: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

01: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

02: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

03: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

04: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

05: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

06: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

07: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

08: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

09: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

10: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

11: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

12: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

13: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

14: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

15: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

16: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

17: 00 00 00 00 00 00 00 00 00 00 12 00 00 00 00 00 00 00 00 00

18: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 12

19: 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

next loop...

0b1331709591d260c1c78e86d0c51c18.png

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