本文目录一览：
- 1、C语言编简单的扫雷
- 2、C语言 扫雷
- 3、C语言扫雷怎么实现?
- 4、C语言扫雷游戏源代码
- 5、扫雷C语言
- 6、如何用C语言编程 扫雷！~
C语言编简单的扫雷
给你一个完整的扫雷源码
#include conio.h
#include graphics.h
#include stdio.h
#include stdlib.h
#include time.h
#include ctype.h
#include "mouse.c"
#define YES 1
#define NO 0
#define XPX 15 /* X pixels by square */
#define YPX 15 /* Y pixels by square */
#define DEFCX 30 /* Default number of squares */
#define DEFCY 28
#define MINE 255-'0' /* So that when it prints, it prints char 0xff */
#define STSQUARE struct stsquare
STSQUARE {
unsigned char value; /* Number of mines in the surround squares */
unsigned char sqopen; /* Square is open */
unsigned char sqpress; /* Square is pressed */
unsigned char sqmark; /* Square is marked */
} *psquare;
#define value(x,y) (psquare+(x)*ncy+(y))-value
#define sqopen(x,y) (psquare+(x)*ncy+(y))-sqopen
#define sqpress(x,y) (psquare+(x)*ncy+(y))-sqpress
#define sqmark(x,y) (psquare+(x)*ncy+(y))-sqmark
int XST, /* Offset of first pixel X */
YST,
ncx, /* Number of squares in X */
ncy,
cmines, /* Mines discovered */
initmines, /* Number of initial mines */
sqclosed, /* Squares still closed */
maxy; /* Max. number of y pixels of the screen */
void Graph_init(void);
void Read_param(int argc, char *argv[]);
void Set_mines(int nminas);
void Set_square(int x, int y, int status);
void Mouse_set(void);
void Draw_squares(void);
int Do_all(void);
void Blow_up(void);
void Open_square(int x, int y);
int Open_near_squares(int x, int y);
/************************************************************************/
void main(int argc, char *argv[])
{
if (!mouse_reset()) {
cputs(" ERROR: I can't find a mouse driver\r\n");
exit(2);
}
Graph_init();
Read_param(argc, argv);
Mouse_set();
do {
randomize();
cleardevice();
Set_mines(cmines=initmines);
mouse_enable();
Draw_squares();
}
while (Do_all() != '\x1b');
closegraph();
}
/*************************************************************************
* *
* F U N C T I O N S *
* *
*************************************************************************/
/*----------------------------------------------------------------------*/
void Graph_init(void)
{
int graphdriver=DETECT, graphmode, errorcode;
if(errorcode 0) {
cprintf("\n\rGraphics System Error: %s\n",grapherrormsg(errorcode));
exit(98);
}
initgraph(graphdriver, graphmode, "");
errorcode=graphresult();
if(errorcode!=grOk) {
printf(" Graphics System Error: %s\n",grapherrormsg(errorcode));
exit(98);
}
maxy=getmaxy();
} /* Graph_init */
/*----------------------------------------------------------------------*/
void Read_param(int argc, char *argv[])
{
int x, y, m;
x=y=m=0;
if (argc!=1) {
if (!isdigit(*argv[1])) {
closegraph();
cprintf("Usage is: %s [x] [y] [m]\r\n\n"
"Where x is the horizontal size\r\n"
" y is the vertical size\r\n"
" m is the number of mines\r\n\n"
" Left mouse button: Open the square\r\n"
"Right mouse button: Mark the square\r\n"
" -The first time puts a 'mine' mark\r\n"
" -The second time puts a 'possible "
"mine' mark\r\n"
" -The third time unmarks the square\r\n"
"Left+Right buttons: Open the surrounded squares only if "
"the count of mines\r\n"
" is equal to the number in the square",argv[0]);
exit (1);
}
switch (argc) {
case 4: m=atoi(argv[3]);
case 3: y=atoi(argv[2]);
case 2: x=atoi(argv[1]);
}
}
XST=100;
ncx=DEFCX;
if (maxy==479) {
YST=30;
ncy=DEFCY;
}
else {
YST=25;
ncy=20;
}
if (x0 xncx)
ncx=x;
if (y0 yncy) {
YST+=((ncy-y)*YPX)1;
ncy=y;
}
initmines= m ? m : ncx*ncy*4/25; /* There are about 16% of mines */
if (((void near*)psquare=calloc(ncx*ncy, sizeof(STSQUARE)))==NULL) {
closegraph();
cputs("ERROR: Not enought memory");
exit(3);
}
} /* Read_param */
/*----------------------------------------------------------------------*/
void Set_mines(int nminas)
{
C语言 扫雷
#includestdio.h
int main(void)
{
char plat[100][100]; //雷的地图
char plat_new[100][100]; //数字映射图
int n, m; //存储行、列数
int in, im;
int mark = 0; //记录该点附近8个坐标雷的总数
int j = 1;
scanf("%d %d", n, m);
getchar(); //消除回车符的影响
do {
if (n == 0 m == 0)
break;
for (in = 0; in n; in++)
{
for (im = 0; im m; im++)
{
scanf("%c", plat[in][im]);
}
getchar();
}
for (in = 0; in n; in++)
for (im = 0; im m; im++)
{
if (plat[in][im] == '*') /*该点有雷，无需检测*/
{
plat_new[in][im]= plat[in][im];
continue;
}
if (in - 1 = 0) //检测上面3个点的雷数
{
if (plat[in - 1][im] == '*')
mark++;
if (im - 1 = 0 plat[in -1][im - 1] == '*')
mark++;
if (im + 1 mplat[in -1][im + 1] == '*')
mark++;
}
if (im - 1 = 0 plat[in][im- 1] == '*') //检测左右两个点的雷数
mark++;
if (im + 1 m plat[in][im+ 1] == '*')
mark++;
if (in + 1 n) //检测下面3个点的雷数
{
if (plat[in + 1][im] == '*')
mark++;
if (im - 1 = 0 plat[in +1][im - 1] == '*')
mark++;
if (im + 1 mplat[in +1][im + 1] == '*')
mark++;
}
switch (mark)
{
case 0:plat_new[in][im] = '0'; break;
case 1:plat_new[in][im] = '1'; break;
case 2:plat_new[in][im] = '2'; break;
case 3:plat_new[in][im] = '3'; break;
case 4:plat_new[in][im] = '4'; break;
case 5:plat_new[in][im] = '5'; break;
case 6:plat_new[in][im] = '6'; break;
case 7:plat_new[in][im] = '7'; break;
case 8:plat_new[in][im] = '8'; break;
}
mark = 0; //重置雷数
}
if (j != 1)
putchar('\n');
printf("Field#%d:\n", j);
for (in = 0; in n; in++) //打印数字地图
{
for (im = 0; im m; im++)
{
printf("%c", plat_new[in][im]);
}
if(in!=n-1)
putchar('\n');
}
scanf("%d %d", n, m);
getchar();
j++;
} while (1);
return 0;
}
C语言扫雷怎么实现?
点击（x,y）
{
如果（x,y）格子不存在，return
如果是炸弹，GG
如果已经翻开，return
如果周围有雷，显示雷数量，并标记翻开
否则
{
显示空，并标记翻开
点击（x-1,y-1）....点击（x+1,y+1）共8个
}
}
基本思路是这样
C语言扫雷游戏源代码
"扫雷"小游戏C代码
#includestdio.h
#includemath.h
#includetime.h
#includestdlib.h
main( )
{char a[102][102],b[102][102],c[102][102],w;
int i,j; /*循环变量*/
int x,y,z[999]; /*雷的位置*/
int t,s; /*标记*/
int m,n,lei; /*计数*/
int u,v; /*输入*/
int hang,lie,ge,mo; /*自定义变量*/
srand((int)time(NULL)); /*启动随机数发生器*/
leb1: /*选择模式*/
printf("\n 请选择模式：\n 1.标准 2.自定义\n");
scanf("%d",mo);
if(mo==2) /*若选择自定义模式，要输入三个参数*/
{do
{t=0; printf("请输入\n行数 列数 雷的个数\n");
scanf("%d%d%d",hang,lie,ge);
if(hang2){printf("行数太少\n"); t=1;}
if(hang100){printf("行数太多\n");t=1;}
if(lie2){printf("列数太少\n");t=1;}
if(lie100){printf("列数太多\n");t=1;}
if(ge1){printf("至少要有一个雷\n");t=1;}
if(ge=(hang*lie)){printf("雷太多了\n");t=1;}
}while(t==1);
}
else{hang=10,lie=10,ge=10;} /*否则就是选择了标准模式(默认参数)*/
for(i=1;i=ge;i=i+1) /*确定雷的位置*/
{do
{t=0; z[i]=rand( )%(hang*lie);
for(j=1;ji;j=j+1){if(z[i]==z[j]) t=1;}
}while(t==1);
}
for(i=0;i=hang+1;i=i+1) /*初始化a,b,c*/
{for(j=0;j=lie+1;j=j+1) {a[i][j]='1'; b[i][j]='1'; c[i][j]='0';} }
for(i=1;i=hang;i=i+1)
{for(j=1;j=lie;j=j+1) {a[i][j]='+';} }
for(i=1;i=ge;i=i+1) /*把雷放入c*/
{x=z[i]/lie+1; y=z[i]%lie+1; c[x][y]='#';}
for(i=1;i=hang;i=i+1) /*计算b中数字*/
{for(j=1;j=lie;j=j+1)
{m=48;
if(c[i-1][j-1]=='#')m=m+1; if(c[i][j-1]=='#')m=m+1;
if(c[i-1][j]=='#')m=m+1; if(c[i+1][j+1]=='#')m=m+1;
if(c[i][j+1]=='#')m=m+1; if(c[i+1][j]=='#')m=m+1;
if(c[i+1][j-1]=='#')m=m+1; if(c[i-1][j+1]=='#')m=m+1;
b[i][j]=m;
}
}
for(i=1;i=ge;i=i+1) /*把雷放入b中*/
{x=z[i]/lie+1; y=z[i]%lie+1; b[x][y]='#';}
lei=ge; /*以下是游戏设计*/
do
{leb2: /*输出*/
system("cls");printf("\n\n\n\n");
printf(" ");
for(i=1;i=lie;i=i+1)
{w=(i-1)/10+48; printf("%c",w);
w=(i-1)%10+48; printf("%c ",w);
}
printf("\n |");
for(i=1;i=lie;i=i+1){printf("---|");}
printf("\n");
for(i=1;i=hang;i=i+1)
{w=(i-1)/10+48; printf("%c",w);
w=(i-1)%10+48; printf("%c |",w);
for(j=1;j=lie;j=j+1)
{if(a[i][j]=='0')printf(" |");
else printf(" %c |",a[i][j]);
}
if(i==2)printf(" 剩余雷个数");
if(i==3)printf(" %d",lei);
printf("\n |");
for(j=1;j=lie;j=j+1){printf("---|");}
printf("\n");
}
scanf("%d%c%d",u,w,v); /*输入*/
u=u+1,v=v+1;
if(w!='#'a[u][v]=='@')
goto leb2;
if(w=='#')
{if(a[u][v]=='+'){a[u][v]='@'; lei=lei-1;}
else if(a[u][v]=='@'){a[u][v]='?'; lei=lei+1;}
else if(a[u][v]=='?'){a[u][v]='+';}
goto leb2;
}
a[u][v]=b[u][v];
leb3: /*打开0区*/
t=0;
if(a[u][v]=='0')
{for(i=1;i=hang;i=i+1)
{for(j=1;j=lie;j=j+1)
{s=0;
if(a[i-1][j-1]=='0')s=1; if(a[i-1][j+1]=='0')s=1;
if(a[i-1][j]=='0')s=1; if(a[i+1][j-1]=='0')s=1;
if(a[i+1][j+1]=='0')s=1; if(a[i+1][j]=='0')s=1;
if(a[i][j-1]=='0')s=1; if(a[i][j+1]=='0')s=1;
if(s==1)a[i][j]=b[i][j];
}
}
for(i=1;i=hang;i=i+1)
{for(j=lie;j=1;j=j-1)
{s=0;
if(a[i-1][j-1]=='0')s=1; if(a[i-1][j+1]=='0')s=1;
if(a[i-1][j]=='0')s=1; if(a[i+1][j-1]=='0')s=1;
if(a[i+1][j+1]=='0')s=1; if(a[i+1][j]=='0')s=1;
if(a[i][j-1]=='0')s=1; if(a[i][j+1]=='0')s=1;
if(s==1)a[i][j]=b[i][j];
}
}
for(i=hang;i=1;i=i-1)
{for(j=1;j=lie;j=j+1)
{s=0;
if(a[i-1][j-1]=='0')s=1; if(a[i-1][j+1]=='0')s=1;
if(a[i-1][j]=='0')s=1; if(a[i+1][j-1]=='0')s=1;
if(a[i+1][j+1]=='0')s=1; if(a[i+1][j]=='0')s=1;
if(a[i][j-1]=='0')s=1; if(a[i][j+1]=='0')s=1;
if(s==1)a[i][j]=b[i][j];
}
}
for(i=hang;i=1;i=i-1)
{for(j=lie;j=1;j=j-1)
{s=0;
if(a[i-1][j-1]=='0')s=1; if(a[i-1][j+1]=='0')s=1;
if(a[i-1][j]=='0')s=1; if(a[i+1][j-1]=='0')s=1;
if(a[i+1][j+1]=='0')s=1;if(a[i+1][j]=='0')s=1;
if(a[i][j-1]=='0')s=1; if(a[i][j+1]=='0')s=1;
if(s==1)a[i][j]=b[i][j];
}
}
for(i=1;i=hang;i=i+1) /*检测0区*/
{for(j=1;j=lie;j=j+1)
{if(a[i][j]=='0')
{if(a[i-1][j-1]=='+'||a[i-1][j-1]=='@'||a[i-1][j-1]=='?')t=1;
if(a[i-1][j+1]=='+'||a[i-1][j+1]=='@'||a[i-1][j+1]=='?')t=1;
if(a[i+1][j-1]=='+'||a[i+1][j-1]=='@'||a[i+1][j-1]=='?')t=1;
if(a[i+1][j+1]=='+'||a[i+1][j+1]=='@'||a[i+1][j+1]=='?')t=1;
if(a[i+1][j]=='+'||a[i+1][j]=='@'||a[i+1][j]=='?')t=1;
if(a[i][j+1]=='+'||a[i][j+1]=='@'||a[i][j+1]=='?')t=1;
if(a[i][j-1]=='+'||a[i][j-1]=='@'||a[i][j-1]=='?')t=1;
if(a[i-1][j]=='+'||a[i-1][j]=='@'||a[i-1][j]=='?')t=1;
}
}
}
if(t==1)goto leb3;
}
n=0; /*检查结束*/
for(i=1;i=hang;i=i+1)
{for(j=1;j=lie;j=j+1)
{if(a[i][j]!='+'a[i][j]!='@'a[i][j]!='?')n=n+1;}
}
}
while(a[u][v]!='#'n!=(hang*lie-ge));
for(i=1;i=ge;i=i+1) /*游戏结束*/
{x=z[i]/lie+1; y=z[i]%lie+1; a[x][y]='#'; }
printf(" ");
for(i=1;i=lie;i=i+1)
{w=(i-1)/10+48; printf("%c",w);
w=(i-1)%10+48; printf("%c ",w);
}
printf("\n |");
for(i=1;i=lie;i=i+1){printf("---|");}
printf("\n");
for(i=1;i=hang;i=i+1)
{w=(i-1)/10+48; printf("%c",w);
w=(i-1)%10+48; printf("%c |",w);
for(j=1;j=lie;j=j+1)
{if(a[i][j]=='0')printf(" |");
else printf(" %c |",a[i][j]);
}
if(i==2)printf(" 剩余雷个数");
if(i==3)printf(" %d",lei); printf("\n |");
for(j=1;j=lie;j=j+1) {printf("---|");}
printf("\n");
}
if(n==(hang*lie-ge)) printf("你成功了!\n");
else printf(" 游戏结束!\n");
printf(" 重玩请输入1\n");
t=0;
scanf("%d",t);
if(t==1)goto leb1;
}
/*注：在DEV c++上运行通过。行号和列号都从0开始，比如要确定第0行第9列不是“雷”，就在0和9中间加入一个字母，可以输入【0a9】三个字符再按回车键。3行7列不是雷，则输入【3a7】回车；第8行第5列是雷，就输入【8#5】回车，9行0列是雷则输入【9#0】并回车*/
扫雷C语言
#include stdio.h
#define N 40
int a[N][2];
int num;
void display()
{
for(int j=0; j num; j++)
{
printf("%d ", a[j][1]);
}
printf("\n");
}
void test(int i)
{
if(i == num)
{
int j;
int flag = 1;
if(a[0][1]+a[1][1]!=a[0][0]a[num-1][1]+a[num-2][1]!=a[num-1][0])
{
}
for(j = 1; j num - 1; j++)
{
if(a[j-1][1] + a[j][1] + a[j+1][1] != a[j][0])
flag = 0;
}
if(flag)
display();
}
for(; i num; i++)
{
if(a[i][1] == 0)
{
if(i == 0)
{
if(a[i][1]+a[i+1][1] != a[i][0])
{
a[i][1] = 1;
test(i+1);
a[i][1] = 0;
}
}
if(i 0)
{
if(a[i-1][1] + a[i][1] + a[i+1][1] != a[i][0])
{
a[i][1] = 1;
test(i+1);
a[i][1] = 0;
}
}
}
}
}
int main()
{
int i;
printf("输入个数：\n");
scanf("%d",num);
printf("输入数据(0~3)：\n");
for(i = 0; i num; i++)
{
scanf("%d",a[i][0]);
a[i][1]=0;
}
for(i = 1; i num - 1; i++)
{
if(a[i][0] == 3)
{
a[i-1][1] = 1;
a[i][1] = 1;
a[i+1][1] = 1;
}
}
test(0);
}
算法思想：
1、如果有输入数字是3则输出数字中对应上中下都必为1
2、输出数组中只有为0的才能为1；
3、用回溯法判断成立条件，成功则输出。
如何用C语言编程 扫雷！~
俄罗斯方快
扫雷
#includestdio.h
#includegraphics.h
#includestdlib.h
struct list
{
int x;
int y;
int num;
int bomb;
int wa;
};
struct list di[10][10];
int currentx=210;
int currenty=130;
void initxy(void)
{
int i,j;
for(i=0;i=9;i++)
for(j=0;j=9;j++)
{
di[j].x=i*20+200;
di[j].y=j*20+120;
di[j].wa=0;
di[j].bomb=0;
}
}
void initmu(void)
{
int i,j;
setcolor(2);
rectangle(200,120,400,320);
rectangle(190,110,410,330);
setfillstyle(8,14);
floodfill(191,111,2);
for(i=0;i=9;i++)
for(j=0;j=9;j++)
rectangle(di[j].x,di[j].y,di[j].x+19,di[j].y+19);
outtextxy(450,200,"press 'enter' to kick");
outtextxy(450,250,"press '\' to mark");
}
void randbomb(void)
{
int k;
int i,j;
randomize();
for(i=0;i=9;i++)
for(j=0;j=9;j++)
{
k=random(5);
if(k==2)
di[j].bomb=1;
}
}
void jisuan(void)
{
int k=0;
int i,j;
for(i=0;i=9;i++)
for(j=0;j=9;j++)
{
if(ijdi[i-1][j-1].bomb)
k=k+1;
if(idi[i-1][j].bomb)
k=k+1;
if(jdi[j-1].bomb)
k=k+1;
if(i=8di[i+1][j].bomb)
k=k+1;
if(j=8di[j+1].bomb)
k=k+1;
if(i=8j=8di[i+1][j+1].bomb)
k=k+1;
if(ij=8di[i-1][j+1].bomb)
k=k+1;
if(i=8jdi[i+1][j-1].bomb)
k=k+1;
di[j].num=k;
k=0;
}
}
void xianbomb(void)
{
int i,j;
char biaoji[2];
char znum[2];
biaoji[0]=1;
biaoji[1]=NULL;
for(i=0;i=9;i++)
for(j=0;j=9;j++)
{
if(di[j].bomb==1)
outtextxy(di[j].x+2,di[j].y+2,biaoji);
else
{
itoa(di[j].num,znum,10);
setfillstyle(1,0);
bar(i*20+202,j*20+122,i*20+218,j*20+138);
outtextxy(i*20+202,j*20+122,znum);
}
}
}
void move(void)
{
int key;
key=bioskey(1);
if(key)
key=bioskey(0);
if(key==0x4800)
{
if(currenty130)
{
setcolor(0);
circle(currentx,currenty,5);
currenty-=20;
setcolor(4);
circle(currentx,currenty,5);
}
else
{
setcolor(0);
circle(currentx,currenty,5);
currenty=310;
setcolor(4);
circle(currentx,currenty,5);
}
}
if(key==0x4b00)
{
if(currentx210)
{
setcolor(0);
circle(currentx,currenty,5);
currentx-=20;
setcolor(4);
circle(currentx,currenty,5);
}
else
{
setcolor(0);
circle(currentx,currenty,5);
currentx=390;
setcolor(4);
circle(currentx,currenty,5);
}
}
if(key==0x4d00)
{
if(currentx390)
{
setcolor(0);
circle(currentx,currenty,5);
currentx+=20;
setcolor(4);
circle(currentx,currenty,5);
}
else
{
setcolor(0);
circle(currentx,currenty,5);
currentx=210;
setcolor(4);
circle(currentx,currenty,5);
}
}
if(key==0x5000)
{
if(currenty310)
{
setcolor(0);
circle(currentx,currenty,5);
currenty+=20;
setcolor(4);
circle(currentx,currenty,5);
}
else
{
setcolor(0);
circle(currentx,currenty,5);
currenty=130;
setcolor(4);
circle(currentx,currenty,5);
}
}
if(key==0x1c0d)
{
int i,j;
char snum[2];
snum[0]=NULL;
snum[1]=NULL;
i=(currentx-210)/20;
j=(currenty-130)/20;
if(di[j].bomb==1)
{
outtextxy(100,100,"game over");
xianbomb();
sleep(2);
exit(0);
}
if(di[j].bomb==0)
{
di[j].wa=1;
setfillstyle(1,0);
bar(currentx-8,currenty-8,currentx+8,currenty+8);
setcolor(15);
itoa(di[j].num,snum,10);
outtextxy(currentx-8,currenty-8,snum);
setcolor(4);
circle(currentx,currenty,5);
}
}
if(key==0x2b5c)
{
char biaoji[2];
biaoji[0]=1;
biaoji[1]=NULL;
setcolor(0);
bar(currentx-8,currenty-8,currentx+8,currenty+8);
setcolor(4);
outtextxy(currentx-8,currenty-8,biaoji);
circle(currentx,currenty,5);
}
}
void success(void)
{
int k=1;
int i,j;
for(i=0;i=9;i++)
for(j=0;j=9;j++)
if(di[j].bomb==0di[j].wa==0)
k=0;
if(k==1)
{
outtextxy(100,100,"success good");
xianbomb();
sleep(2);
exit(0);
}
}
void main(void)
{
int gd=DETECT,gm;
initgraph(gd,gm,"");
initxy();
initmu();
randbomb();
jisuan();
setcolor(4);
circle(210,130,5);
while(1)
{
move();
success();
}
}
