整体架构如上
1.很直观地去看这个中国象棋的界面,数一下它有多少行和多少列.
10行,9列:要注意这里数的是安放象棋的位置,有10行9列
这里我们首先想到的必然是二维数组,每一个行列交叉的点都设置成二维数组a[i][j]这样的格式,以此来确定棋盘上面每一个棋子的位置和走向.
我们把上面安放棋子的二维数组定义成一个地图,也就是map[i][j]的格式,但同时要注意,每一个象棋棋子都有行,列,颜色,是否过河和名称(也就是什么棋子)几种定义区分.
那么这里我们就需要把棋子定义成一个结构.如下:
struct Chesscoordinate//棋子综合信息
{int x;int y;DWORD type; //颜色bool river;//是否过河int id;
};
当定义了棋子综合信息,我们是不是需要每一个信息都拓展一下?
x代表的是在棋盘上面的列,也就是竖行,y代表的是棋子在棋盘上面的行,也就是横线,而type代表的棋子的颜色,棋子颜色可以分为黑色和红色两种;river设置成bool型的,只需要判断棋子是否过河就可以了.最后id定义的是棋子上面的名称,比如将,帅等.
接下来我们就来VS当中进行棋子的程序定义拓展:
#define distance 35//窗口线与棋盘边界线的距离
#define longth 65//棋盘方格的长
#define high 61//棋盘方格的宽
distance longth 和high我们都把其设置成宏,这个就要回溯到棋盘上面,把棋盘假设成一个xy的二维坐标,那么要定义每一个棋子的位置,或者说是每一行与每一列的交叉点,就要用到上面三个宏定义.
比如左上角第一个棋子的坐标可以表示为:
map[0][0].x=distance;
map[0][0].y=distance
而第一行第二个棋子的坐标可以表示为:
map[1][0].x=distance+longth;
map[1][0].y=distance
依次类推......
int xx(int a)//数组下标转换成坐标
{a = distance + longth * a;return a;
}
int yy(int a)
{a = distance + high * a;return a;
}
我们可以推断出每一个棋子的坐标,都可以通过上面几个宏定义以及棋子的实际行列表示出来,
enum pieces
{SPACE = -1,車, 馬, 象, 士, 将, 炮, 卒,车, 马, 相, 仕, 帥, 砲, 兵,BEGIN, END,};
enum pieces redchess[] = { 車,馬,象,士,将,炮,卒, };
enum pieces blackchess[] = { 车,马,相,仕,帥,砲,兵, };
enum pieces state = BEGIN;
struct move { //鼠标选中的棋子int beginx;int beginy;int endx;int endy;int state;
}moving = { -1,-1,-1,-1,BEGIN };
const char* chessname[] = { "車","馬","象","士","将","炮","卒","车","马","相","仕","帥","砲","兵", };
这里我们把棋子的id设置成一个联合结构数组,因为里面的棋子的id都基本不同,同时我们考虑到了棋子要进行运动,把棋子的开始状态设置为BEGIN
而新建一个move的结构,表示鼠标选中棋子的基本信息.begin的行列和结束的行列和运行状态.
最后回到棋子id上面,我们是不是需要把这些id都设置到棋子或者棋盘上面啊,采用常量字符数组的形式,把棋盘上面所有的棋子都表示出来.
而把刚才的map[i][j]二维数组同样定义成一个二维结构数组.如下:
struct Chesscoordinate map[9][10];//坐标
struct Chesscoordinate AImap[9][10];//坐标
同时,我们设置了AImap[i][j]的二维结构数组,看看是否下面在进行棋子移动的时候,我们会用到.
这里i=9,j=10是根据我们上面看中国象棋棋盘得到的数据.
2.我们要开始加载图片信息了,把我们已经有的棋盘素材放在同文件下面,然后采用下面程序,就可以在窗口当中显示出中国棋盘的背景图.
void begining()
{loadimage(&img, "D:/program/Project3-chess successful/1.png");initgraph(img.getwidth(), img.getheight(), 1); //初始化图形系统putimage(0, 0, &img);//输出开始界面;
至于,loadimage()和initgraph()函数以及putimage()函数三者的基本理解和解释,建议去搜索,都有比较详细的解释.这里我们解释一下initgraph()函数的基本含义:
这个函数用于初始化绘图环境。
HWND initgraph(int width,int height,int flag = NULL
);
参数:
width 绘图环境的宽度。
height 绘图环境的高度。
flag 绘图环境的样式,默认为 NULL。
上面程序的基本思路也就不言而喻:先加载或者找到我们素材的地址,然后定义出素材的宽和长度,然后再输出这个棋盘背景.
当输出完背景之后,我们是不是要先把所有的棋子放在初始位置,也就是初始化棋盘?
void coord()//棋子信息赋值
{loadimage(&img, "D:/program/Project3-chess successful/1.png");putimage(0, 0, &img);//输出棋盘for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++)//遍历二维数组{enum pieces chessid = SPACE;//先把全部位置的id赋值为SAPCEDWORD chesstype;//定义颜色中间变量if (j <= 4){chesstype = BLACK;//黑方if (j == 0){if (i <= 4){chessid = blackchess[i];//}else{chessid = blackchess[8 - i];}}if (j == 2 && (i == 1 || i == 7)){chessid = blackchess[5];}if (j == 3 && (i == 0 || i == 2 || i == 4 || i == 4 || i == 6 || i == 8)){chessid = blackchess[6];}}else{chesstype = RED;//红方if (j == 6 && (i == 0 || i == 2 || i == 4 || i == 6 || i == 8)){chessid = redchess[6];}if (j == 7 && (i == 1 || i == 7)){chessid = redchess[5];}if (j == 9){if (i <= 4){chessid = redchess[i];}else{chessid = redchess[8 - i];}}}//依次赋值map[i][j].id = chessid;map[i][j].river = false;map[i][j].type = chesstype;map[i][j].x = distance + longth * i;map[i][j].y = distance + high * j;}}for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id == SPACE){map[i][j].type = YELLOW;}}}}
看上述程序:
首先先加载出棋盘背景,然后把棋盘上面行和列交叉的位置都设置为SPACE,可以理解为初始化棋盘.然后如果j<=4,也就是楚河的一侧,我们定义棋子type全部都为黑色,而棋子的id要根据行列的位置来确定,首先看第一行,前5列数据,是不是应该是"車,馬,象,士,将"而后四列数据,是不是应该是前面去除将的反序,而观察一下,如果把前面定义成i列,后面对应相等的字符则是8-i列.那么我们就定义完了第一行的数据.
而炮和兵的数据,以及红方的数据是不是同上述思路一样.
然后就开始定义每一个棋子初始化的状态,回溯到棋子的基本信息结构当中,有x,y,river,type和id,我们要依次对其进行定义,
注意在初始状态的时候,棋子的river都为false,因为都没有过楚河
定义完棋子的位置,那么剩下没有棋子的位置,我们是不是要把刚开始初始化的状态都修改为棋盘的背景颜色啊.上述定义完,再次输出棋盘
void qiban()
{loadimage(&img, "D:/program/Project3-chess successful/1.png");initgraph(img.getwidth(), img.getheight(), 1);putimage(0, 0, &img);//输出棋盘
}
这个时候就已经初始化了中国象棋的棋盘.
接下来是不是要把每一个棋子都定义到棋盘上面.开始绘画棋子
void getbackground() //画棋子
{int x_start, y_start;x_start = distance;y_start = distance;settextstyle(30, 0, "黑体");//棋子大小颜色setbkmode(0);for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id != SPACE)//在棋盘上输出{setfillcolor(RGB(253, 216, 161));// setlinestyle(BLACK);settextcolor(map[i][j].type);fillcircle(map[i][j].x, map[i][j].y, 24);fillcircle(map[i][j].x, map[i][j].y, 18);outtextxy(map[i][j].x - 13, map[i][j].y - 13, chessname[map[i][j].id]);}}}}
如果上面棋盘行列之间的id不是SPACE,那自然就是我们定义的棋子应该安置的位置.从第0行第0列开始,依次进行定义:
画一个棋子,两个圆环,其颜色要画成背景颜色,而棋子上面的文本颜色要根据实际行列定义的type,设置成黑色和红色,
最后便是采用outtextxy()把每一个棋子的字符输入进行.注意此处根据棋盘x和y的地址均减13,是根据实际调试来进行,并不是确定常量.
写到这里,是不是就把棋子和棋盘都初始化好了?yes,of course.
3.棋盘棋子都初始化好了,那么接下来我们是不是就可以开始移动棋盘上面的棋子,进行下棋了?
这里有一个有意思的思考:我刚开始认为吃掉棋子和走到空白的地方是两种不同的情况,后面仔细思考了一下,是同一状态,只需要把原来的修改为结束的就可以了,而原来的只需要位置只需要修改颜色和id为空也就没了,不管他end地址有没有棋子.
void movechess(int a, int b, int c, int d)//移动棋子,改变其坐标
{map[c][d].id = map[a][b].id;map[c][d].river = map[a][b].river;map[c][d].type = map[a][b].type;map[c][d].x = xx(c);map[c][d].y = yy(d);map[a][b].id = SPACE;map[a][b].type = YELLOW;}
int xx(int a)//数组下标转换成坐标
{a = distance + longth * a;return a;
}
int yy(int a)
{a = distance + high * a;return a;
}
把[a][b]位置的棋子信息,全部给了[c][d]位置处的棋子,而[a][b]棋子此处的信息要记得修改id和type,id设置为空,而颜色设置成背景颜色.而xx(c),yy(d)就成了棋子移动到下一个点的坐标
void MouseControl()//获取鼠标点击信息并完响应
{//getbackground();if (MouseHit()){float beginrow, beginrol, endrow, endrol;//第一次按下时的坐标int intbeginrow, intbeginrol, intendrow, intendrol;//第二次按下时的坐标MOUSEMSG msg = GetMouseMsg();/*这个函数用于获取一个鼠标消息。如果当前鼠标消息队列中没有,就一直等待。*/if (msg.uMsg == WM_LBUTTONDOWN)//按下鼠标左键时{//获取鼠标点击的数组的下标// printf("(%d,%d)", msg.x, msg.y);//回溯转换成行列beginrow = (float)(msg.x - distance) / longth;beginrol = (float)(msg.y - distance) / high;intbeginrow = round(beginrow);intbeginrol = round(beginrol);if (moving.state == BEGIN){moving.state = END;moving.beginx = intbeginrow;moving.beginy = intbeginrol;// printf("(%d,%d) \n", moving.beginx, moving.beginy);}else if (moving.state == END){moving.state = BEGIN;moving.endx = intbeginrow;moving.endy = intbeginrol;execute(moving.beginx, moving.beginy, moving.endx, moving.endy);}}}}
上面这一段我其实有一点不清晰,我所理解的应该是定义了两个状态,如果状态是开始的时候,那状态改为结束,然后把第一次点击的数据给了开始的数据,如果状态是结束,把状态修改为开始,把第一次点击的数据给了结束的坐标.
关键:
intbeginrow = round(beginrow);intbeginrol = round(beginrol);
判断输赢:
int win()
{int redgeneral = 0;int blackgeneral = 0;for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id == blackchess[4]){blackgeneral++;}else if (map[i][j].id == redchess[4]){redgeneral++;}else{blackgeneral = blackgeneral;redgeneral = redgeneral;}}}//printf("%d %d\n", blackgeneral, redgeneral);if (blackgeneral == 0)//红方胜{return 0;}else if (redgeneral == 0)//黑方胜{return 1;}else//打平{return 2;}}
给将和帅都定义了初始值,相当于如果将死了,那就是帅赢,相反同理,而如果两者都存在,也就是1+1,那自然就是平手,return 2;
下面就是判断每一种棋子是否按游戏规则进行移动:
bool jiang(int a, int b, int c, int d)//判断是否只移动了一格(将军、兵的规则)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (b < 4 && c > 2 && c < 6 && d < 3){if (h == 1 && map[c][d].type != map[a][b].type)return true;elsereturn false;}if (b > 4 && c > 2 && c < 6 && d >6){if (h == 1 && map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}
bool bing(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (map[a][b].type == BLACK){if (map[a][b].river == false){if (d == b + 1 && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}else{if (d >= b && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}}else if (map[a][b].type == RED){if (map[a][b].river == false){if (d == b - 1 && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}else{if (d <= b && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}}else{return false;}
}
bool pao(int a, int b, int c, int d)//炮的移动
{if (c == a && d != b){int time = 0;int max = d > b ? d : b;int min = b < d ? b : d;for (int i = min; i <= max; i++){if (map[c][i].id != SPACE){time++;}}// printf("%d\n", time);if (map[c][d].id == SPACE){if (time == 1)return true;elsereturn false;}if (map[c][d].id != SPACE){if (time != 3){return false;}else{if (map[c][d].type == map[a][b].type){return false;}else{return true;}}}}else if (d == b && c != a){int time = 0;int max = a > c ? a : c;int min = c < a ? c : a;for (int i = min; i <= max; i++){if (map[i][d].id != SPACE){time++;}}// printf("%d\n", time);if (map[c][d].id == SPACE){if (time == 1)return true;elsereturn false;}if (map[c][d].id != SPACE){if (time != 3){return false;}else{if (map[c][d].type == map[a][b].type){return false;}else{return true;}}}}else{return false;}
}
bool che(int a, int b, int c, int d)
{if (c == a && d != b)//是否为直线{int time = 0;int max = d > b ? d : b;int min = b < d ? b : d;for (int i = min; i <= max; i++)//遍历路径{if (map[c][i].id != SPACE){time++;}}// printf("%d", time);if (time == 1)//车移动不吃棋子{return true;}if (time == 2)//车移动并且吃目的坐标的棋子{if (map[c][d].type == map[a][b].type)//如果是目的坐标是自己的棋子,则返回false{return false;}if (map[c][d].type == YELLOW){return false;}else{return true;}}else{return false;}}else if (d == b && c != a){int time = 0;int max = c > a ? c : a;int min = a < c ? a : c;for (int i = min; i <= max; i++)//遍历路径{if (map[i][d].id != SPACE){time++;}}// printf("%d", time);if (time == 1)//车是否车跳棋{return true;}else if (time == 2){if (map[c][d].type == map[a][b].type){return false;}if (map[c][d].type == YELLOW){return false;}else{return true;}}else{return false;}}else{return 0;}
}
bool ma(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));// printf("%f", h);if (h <= 2 || h >= 2.5)//根号8=2.8.根号5=2.2{// printf("太远了!\n");return false;}else{int xx, yy, max, min;//关键点的坐标和中间值max = abs(d - b) > abs(c - a) ? abs(d - b) : abs(c - a);min = abs(c - a) < abs(d - b) ? abs(c - a) : abs(d - b);//printf("max\min:(%d,%d)", max, min);if (max == abs(d - b)){yy = b + (d - b) / 2;xx = a;}else{xx = a + (c - a) / 2;yy = b;}// printf("xx\yy:(%d,%d)\n", xx, yy);if (map[xx][yy].id == SPACE){if (map[c][d].type != map[a][b].type){// printf("目的坐标(%d,%d)\n", c, d);// printf("那是你自己的棋子!\n");return true;}else{// printf("那是你的棋子!\n");return false;}}else{// printf("关键位置有棋子!\n");return false;}}
}
bool xiang(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (b <= 4){if (d > 4){return false;}else{if (h < 2.4 || h>2.9){return false;}else{int xx = (a + c) / 2;int yy = (b + d) / 2;if (map[xx][yy].id == SPACE){if (map[c][d].type == map[a][b].type){return false;}else{return true;}}else{return false;}}}}else{if (d < 5){return false;}else{if (h < 2.4 || h>2.9){return false;}else{int xx = (a + c) / 2;int yy = (b + d) / 2;if (map[xx][yy].id == SPACE){if (map[c][d].type == map[a][b].type){return false;}else{return true;}}else{return false;}}}}}
bool shi(int a, int b, int c, int d)
{float h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));// printf("%f", h)if (b < 5){if (c >= 3 && c <= 5 && d <= 2){if (1.2 < h && h < 1.5){if (map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}else{return false;}}else if (b > 5){if (c >= 3 && c <= 5 && d >= 7){if (1.2 < h && h < 1.5){if (map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}else{return false;}}elsereturn false;
}
上述全都是bool型,用以表示判断是否正确就行
接下来是行棋子,如果遵守规则,也就是上面的bool输出true,则移动,否则输出"你不能这样做!"
void execute(int a, int b, int c, int d)//行棋
{if (map[a][b].id == blackchess[4])//黑方将{if (jiang(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做\n");}}else if (map[a][b].id == redchess[4])//红方将{if (jiang(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[6])//黑方兵{if (map[a][b].river == false){if (bing(a, b, c, d)){movechess(a, b, c, d);if (d > 4){map[c][d].river = true;}}else{printf("你不可以这样做!\n");}}else{if (bing(a, b, c, d) && d >= b){movechess(a, b, c, d);}else{printf("你不可以这样做\n");}}}else if (map[a][b].id == redchess[6])//红方兵{if (map[a][b].river == false){if (bing(a, b, c, d)){movechess(a, b, c, d);if (d < 5){map[c][d].river = true;}}else{printf("你不可以这样做!\n");}}else{if (bing(a, b, c, d) && d <= b){movechess(a, b, c, d);}else{printf("你不可以这样做!\n");}}}else if (map[a][b].id == blackchess[5] || map[a][b].id == redchess[5]){if (pao(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[0] || map[a][b].id == redchess[0]){if (che(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不可以这样做!\n");}}else if (map[a][b].id == blackchess[1] || map[a][b].id == redchess[1]){if (ma(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[2] || map[a][b].id == redchess[2]){if (xiang(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!\n");}else if (map[a][b].id == blackchess[3]){if (shi(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!");}else if (map[a][b].id == redchess[3]){if (shi(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!");}
}
定义完上面所有的函数,最后就是主函数,运行就可以了
int main()
{begining();while (1)//当正确的时候{coord();//输出棋盘win();BeginBatchDraw();/*这个函数用于开始批量绘图。执行后任何绘图操作都将暂时不输出到绘图窗口上,直到执行 FlushBatchDraw 或 EndBatchDraw 才将之前的绘图输出。*///运行BeginBatchDraw后,所有的绘图都不再显示在屏幕上,而是在内存中进行//封装的双缓存,避免闪屏/*那如果我们能让打印的过程不显示出来,只显示打印完后的显示缓冲区不就行了吗*/while (win() == 2)//平局状态,是不是还需要继续运行下去?{win();putimage(0, 0, &img);getbackground();//输出棋子MouseControl();//鼠标更改数据FlushBatchDraw();/*这个函数用于执行未完成的绘制任务。*/}putimage(0, 0, &img);getbackground();//输出棋子MouseControl();//鼠标更改数据FlushBatchDraw();/*这个函数用于执行未完成的绘制任务。*/if (win() == 0){printf("红方胜!\n");}else if (win() == 1){printf("黑方胜!\n");}}getchar();return 0;
}
总代码:
#include<stdio.h>
#include<graphics.h>
#include<math.h>
void execute(int a, int b, int c, int d);
bool jiang(int a, int b, int c, int d);
bool pao(int a, int b, int c, int d);
bool ma(int a, int b, int c, int d);
IMAGE img;
#define distance 35//窗口线与棋盘边界线的距离
#define longth 65//棋盘方格的长
#define high 61//棋盘方格的宽
struct movecoordinate
{long x;long y;
};
struct Chesscoordinate//棋子综合信息
{int x;int y;DWORD type; //颜色bool river;//是否过河int id;
};
enum pieces
{SPACE = -1,車, 馬, 象, 士, 将, 炮, 卒,车, 马, 相, 仕, 帥, 砲, 兵,BEGIN, END,};
enum pieces redchess[] = { 車,馬,象,士,将,炮,卒, };
enum pieces blackchess[] = { 车,马,相,仕,帥,砲,兵, };
enum pieces state = BEGIN;
struct move { //鼠标选中的棋子int beginx;int beginy;int endx;int endy;int state;
}moving = { -1,-1,-1,-1,BEGIN };
const char* chessname[] = { "車","馬","象","士","将","炮","卒","车","马","相","仕","帥","砲","兵", };
struct Chesscoordinate map[9][10];//坐标
struct Chesscoordinate AImap[9][10];//坐标
movecoordinate begin = { -1,-1 }, end = { -1,-1 };
int xx(int a)//数组下标转换成坐标
{a = distance + longth * a;return a;
}
int yy(int a)
{a = distance + high * a;return a;
}
void begining()
{loadimage(&img, "D:/program/Project3-chess successful/1.png");initgraph(img.getwidth(), img.getheight(), 1); //初始化图形系统putimage(0, 0, &img);//输出开始界面;}
void qiban()
{loadimage(&img, "D:/program/Project3-chess successful/1.png");initgraph(img.getwidth(), img.getheight(), 1);putimage(0, 0, &img);//输出棋盘
}
void coord()//棋子信息赋值
{loadimage(&img, "D:/program/Project3-chess successful/1.png");putimage(0, 0, &img);//输出棋盘for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++)//遍历二维数组{enum pieces chessid = SPACE;//先把全部位置的id赋值为SAPCEDWORD chesstype;//定义颜色中间变量if (j <= 4){chesstype = BLACK;//黑方if (j == 0){if (i <= 4){chessid = blackchess[i];//}else{chessid = blackchess[8 - i];}}if (j == 2 && (i == 1 || i == 7)){chessid = blackchess[5];}if (j == 3 && (i == 0 || i == 2 || i == 4 || i == 4 || i == 6 || i == 8)){chessid = blackchess[6];}}else{chesstype = RED;//红方if (j == 6 && (i == 0 || i == 2 || i == 4 || i == 6 || i == 8)){chessid = redchess[6];}if (j == 7 && (i == 1 || i == 7)){chessid = redchess[5];}if (j == 9){if (i <= 4){chessid = redchess[i];}else{chessid = redchess[8 - i];}}}//依次赋值map[i][j].id = chessid;map[i][j].river = false;map[i][j].type = chesstype;map[i][j].x = distance + longth * i;map[i][j].y = distance + high * j;}}for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id == SPACE){map[i][j].type = YELLOW;}}}}
void getbackground() //画棋子
{int x_start, y_start;x_start = distance;y_start = distance;settextstyle(30, 0, "黑体");//棋子大小颜色setbkmode(0);for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id != SPACE)//在棋盘上输出{setfillcolor(RGB(253, 216, 161));// setlinestyle(BLACK);settextcolor(map[i][j].type);fillcircle(map[i][j].x, map[i][j].y, 24);fillcircle(map[i][j].x, map[i][j].y, 18);outtextxy(map[i][j].x - 13, map[i][j].y - 13, chessname[map[i][j].id]);}}}}
void movechess(int a, int b, int c, int d)//移动棋子,改变其坐标
{map[c][d].id = map[a][b].id;map[c][d].river = map[a][b].river;map[c][d].type = map[a][b].type;map[c][d].x = xx(c);map[c][d].y = yy(d);map[a][b].id = SPACE;map[a][b].type = YELLOW;}void MouseControl()//获取鼠标点击信息并完响应
{//getbackground();if (MouseHit()){float beginrow, beginrol, endrow, endrol;//第一次按下时的坐标int intbeginrow, intbeginrol, intendrow, intendrol;//第二次按下时的坐标MOUSEMSG msg = GetMouseMsg();/*这个函数用于获取一个鼠标消息。如果当前鼠标消息队列中没有,就一直等待。*/if (msg.uMsg == WM_LBUTTONDOWN)//按下鼠标左键时{//获取鼠标点击的数组的下标// printf("(%d,%d)", msg.x, msg.y);//回溯转换成行列beginrow = (float)(msg.x - distance) / longth;beginrol = (float)(msg.y - distance) / high;intbeginrow = round(beginrow);intbeginrol = round(beginrol);if (moving.state == BEGIN){moving.state = END;moving.beginx = intbeginrow;moving.beginy = intbeginrol;// printf("(%d,%d) \n", moving.beginx, moving.beginy);}else if (moving.state == END){moving.state = BEGIN;moving.endx = intbeginrow;moving.endy = intbeginrol;execute(moving.beginx, moving.beginy, moving.endx, moving.endy);}}}}int win()
{int redgeneral = 0;int blackgeneral = 0;for (int i = 0; i <= 8; i++){for (int j = 0; j <= 9; j++){if (map[i][j].id == blackchess[4]){blackgeneral++;}else if (map[i][j].id == redchess[4]){redgeneral++;}else{blackgeneral = blackgeneral;redgeneral = redgeneral;}}}//printf("%d %d\n", blackgeneral, redgeneral);if (blackgeneral == 0)//红方胜{return 0;}else if (redgeneral == 0)//黑方胜{return 1;}else//打平{return 2;}}
bool jiang(int a, int b, int c, int d)//判断是否只移动了一格(将军、兵的规则)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (b < 4 && c > 2 && c < 6 && d < 3){if (h == 1 && map[c][d].type != map[a][b].type)return true;elsereturn false;}if (b > 4 && c > 2 && c < 6 && d >6){if (h == 1 && map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}
bool bing(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (map[a][b].type == BLACK){if (map[a][b].river == false){if (d == b + 1 && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}else{if (d >= b && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}}else if (map[a][b].type == RED){if (map[a][b].river == false){if (d == b - 1 && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}else{if (d <= b && h == 1 && map[c][d].type != map[a][b].type){return true;}else{return false;}}}else{return false;}
}
bool pao(int a, int b, int c, int d)//炮的移动
{if (c == a && d != b){int time = 0;int max = d > b ? d : b;int min = b < d ? b : d;for (int i = min; i <= max; i++){if (map[c][i].id != SPACE){time++;}}// printf("%d\n", time);if (map[c][d].id == SPACE){if (time == 1)return true;elsereturn false;}if (map[c][d].id != SPACE){if (time != 3){return false;}else{if (map[c][d].type == map[a][b].type){return false;}else{return true;}}}}else if (d == b && c != a){int time = 0;int max = a > c ? a : c;int min = c < a ? c : a;for (int i = min; i <= max; i++){if (map[i][d].id != SPACE){time++;}}// printf("%d\n", time);if (map[c][d].id == SPACE){if (time == 1)return true;elsereturn false;}if (map[c][d].id != SPACE){if (time != 3){return false;}else{if (map[c][d].type == map[a][b].type){return false;}else{return true;}}}}else{return false;}
}
bool che(int a, int b, int c, int d)
{if (c == a && d != b)//是否为直线{int time = 0;int max = d > b ? d : b;int min = b < d ? b : d;for (int i = min; i <= max; i++)//遍历路径{if (map[c][i].id != SPACE){time++;}}// printf("%d", time);if (time == 1)//车移动不吃棋子{return true;}if (time == 2)//车移动并且吃目的坐标的棋子{if (map[c][d].type == map[a][b].type)//如果是目的坐标是自己的棋子,则返回false{return false;}if (map[c][d].type == YELLOW){return false;}else{return true;}}else{return false;}}else if (d == b && c != a){int time = 0;int max = c > a ? c : a;int min = a < c ? a : c;for (int i = min; i <= max; i++)//遍历路径{if (map[i][d].id != SPACE){time++;}}// printf("%d", time);if (time == 1)//车是否车跳棋{return true;}else if (time == 2){if (map[c][d].type == map[a][b].type){return false;}if (map[c][d].type == YELLOW){return false;}else{return true;}}else{return false;}}else{return 0;}
}
bool ma(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));// printf("%f", h);if (h <= 2 || h >= 2.5)//根号8=2.8.根号5=2.2{// printf("太远了!\n");return false;}else{int xx, yy, max, min;//关键点的坐标和中间值max = abs(d - b) > abs(c - a) ? abs(d - b) : abs(c - a);min = abs(c - a) < abs(d - b) ? abs(c - a) : abs(d - b);//printf("max\min:(%d,%d)", max, min);if (max == abs(d - b)){yy = b + (d - b) / 2;xx = a;}else{xx = a + (c - a) / 2;yy = b;}// printf("xx\yy:(%d,%d)\n", xx, yy);if (map[xx][yy].id == SPACE){if (map[c][d].type != map[a][b].type){// printf("目的坐标(%d,%d)\n", c, d);// printf("那是你自己的棋子!\n");return true;}else{// printf("那是你的棋子!\n");return false;}}else{// printf("关键位置有棋子!\n");return false;}}
}
bool xiang(int a, int b, int c, int d)
{float h;h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));if (b <= 4){if (d > 4){return false;}else{if (h < 2.4 || h>2.9){return false;}else{int xx = (a + c) / 2;int yy = (b + d) / 2;if (map[xx][yy].id == SPACE){if (map[c][d].type == map[a][b].type){return false;}else{return true;}}else{return false;}}}}else{if (d < 5){return false;}else{if (h < 2.4 || h>2.9){return false;}else{int xx = (a + c) / 2;int yy = (b + d) / 2;if (map[xx][yy].id == SPACE){if (map[c][d].type == map[a][b].type){return false;}else{return true;}}else{return false;}}}}}
bool shi(int a, int b, int c, int d)
{float h = sqrt(abs(d - b) * abs(d - b) + abs(c - a) * abs(c - a));// printf("%f", h)if (b < 5){if (c >= 3 && c <= 5 && d <= 2){if (1.2 < h && h < 1.5){if (map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}else{return false;}}else if (b > 5){if (c >= 3 && c <= 5 && d >= 7){if (1.2 < h && h < 1.5){if (map[c][d].type != map[a][b].type)return true;elsereturn false;}else{return false;}}else{return false;}}elsereturn false;
}
void execute(int a, int b, int c, int d)//行棋
{if (map[a][b].id == blackchess[4])//黑方将{if (jiang(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做\n");}}else if (map[a][b].id == redchess[4])//红方将{if (jiang(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[6])//黑方兵{if (map[a][b].river == false){if (bing(a, b, c, d)){movechess(a, b, c, d);if (d > 4){map[c][d].river = true;}}else{printf("你不可以这样做!\n");}}else{if (bing(a, b, c, d) && d >= b){movechess(a, b, c, d);}else{printf("你不可以这样做\n");}}}else if (map[a][b].id == redchess[6])//红方兵{if (map[a][b].river == false){if (bing(a, b, c, d)){movechess(a, b, c, d);if (d < 5){map[c][d].river = true;}}else{printf("你不可以这样做!\n");}}else{if (bing(a, b, c, d) && d <= b){movechess(a, b, c, d);}else{printf("你不可以这样做!\n");}}}else if (map[a][b].id == blackchess[5] || map[a][b].id == redchess[5]){if (pao(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[0] || map[a][b].id == redchess[0]){if (che(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不可以这样做!\n");}}else if (map[a][b].id == blackchess[1] || map[a][b].id == redchess[1]){if (ma(a, b, c, d)){movechess(a, b, c, d);}else{printf("你不能这样做!\n");}}else if (map[a][b].id == blackchess[2] || map[a][b].id == redchess[2]){if (xiang(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!\n");}else if (map[a][b].id == blackchess[3]){if (shi(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!");}else if (map[a][b].id == redchess[3]){if (shi(a, b, c, d)){movechess(a, b, c, d);}elseprintf("你不能这样做!");}
}
int main()
{begining();while (1)//当正确的时候{coord();//输出棋盘win();BeginBatchDraw();/*这个函数用于开始批量绘图。执行后任何绘图操作都将暂时不输出到绘图窗口上,直到执行 FlushBatchDraw 或 EndBatchDraw 才将之前的绘图输出。*///运行BeginBatchDraw后,所有的绘图都不再显示在屏幕上,而是在内存中进行//封装的双缓存,避免闪屏/*那如果我们能让打印的过程不显示出来,只显示打印完后的显示缓冲区不就行了吗*/while (win() == 2)//平局状态,是不是还需要继续运行下去?{win();putimage(0, 0, &img);getbackground();//输出棋子MouseControl();//鼠标更改数据FlushBatchDraw();/*这个函数用于执行未完成的绘制任务。*/}putimage(0, 0, &img);getbackground();//输出棋子MouseControl();//鼠标更改数据FlushBatchDraw();/*这个函数用于执行未完成的绘制任务。*/if (win() == 0){printf("红方胜!\n");}else if (win() == 1){printf("黑方胜!\n");}}getchar();return 0;
}
总代码参考于:
(1条消息) 基于c语言的象棋游戏_m0_59564114的博客-CSDN博客_c语言象棋游戏代码