• 《连连看》算法c语言演示(自动连连看)



    (图片是游戏的示意图,来自互联网,与本文程序无关)

    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:

    1. 两个图片相同。
    2. 两个图片之间,沿着相邻的格子画线,中间不能有障碍物。
    3. 画线中间最多允许2个转折。

    所以算法主要是这样几部分:

    1. 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。
    2. 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。
    3. 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。
    4. 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。
    5. 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。
    6. 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<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<time.h>
    
    //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
    ",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;i++){
            setRandPic();
        }
        return;
    }
    //-----------------------------show status --------------------
    //显示当前的图板情况
    void dumpMap(){
        int i,j;
        printf("--: ");
        for(i=0;i<_width;i++){
            printf("%02d ",i);
        }
        printf("
    ");
        for(i=0;i<_height;i++){
            printf("%02d: ",i);
            for(j=0;j<_width;j++){
                printf("%02d ",map[j][i]);
            }
            printf("
    ");
        }
    }
    //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;x++){
            printf("%02d ",x);
        }
        printf("
    ");
        for(y=0;y<_height;y++){
            printf("%02d: ",y);
            for(x=0;x<_width;x++){
                if ((c1.x==x && c1.y==y) || (c2.x==x && c2.y==y))
                    printf("e[1;31m%02de[0m ",map[x][y]);
                else
                    printf("%02d ",map[x][y]);
            }
            printf("
    ");
        }
    }
    //-------------------------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;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 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;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;
        }
        //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.
    ",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].
    ",
                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].
    ",
                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
    ",p1.x,p1.y);
        int c1=map[p1.x][p1.y];
        for (iy=p1.y;iy<_height;iy++){
            for(ix=0;ix<_width;ix++){
                //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
    ",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]
    ",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;iy++){
                for(ix=0;ix<_width;ix++){
                    if(map[ix][iy] != empty){
                        _point p;
                        p.x=ix;
                        p.y=iy;
                        if(searchMap(p) && noPathLeft)
                            noPathLeft=FALSE;
                    }
                }
            }
            printf("next loop...
    ");
        };
        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...
    
    
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  • 原文地址:https://www.cnblogs.com/andrewwang/p/9181874.html
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