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C语言扫雷函数流程图,c语言扫雷程序设计
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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)
{
相关问答
Q1: 要个扫雷的c语言算法
先把你上次问的围棋的给你，这个扫雷的我再写
*主函数*/ void main() { int press; int bOutWhile=FALSE;/*退出循环标志*/
Init();/*初始化图象，数据*/
while(1) { press=GetKey();/*获取用户的按键值*/ switch(CheckKey(press))/*判断按键类别*/ { /*是退出键*/ case KEYEXIT: clrscr();/*清屏*/ bOutWhile = TRUE; break;
/*是落子键*/ case KEYFALLCHESS: if(ChessGo(gPlayOrder,gCursor)==FALSE)/*走棋*/ DoError();/*落子错误*/ else { DoOK();/*落子正确*/
/*如果当前行棋方赢棋*/ if(JudgeWin(gPlayOrder,gCursor)==TRUE) { DoWin(gPlayOrder); bOutWhile = TRUE;/*退出循环标志置为真*/ } /*否则*/ else /*交换行棋方*/ ChangeOrder(); } break;
/*是光标移动键*/ case KEYMOVECURSOR: MoveCursor(gPlayOrder,press); break;
/*是无效键*/ case KEYINVALID: break; }
if(bOutWhile==TRUE) break; }
/*游戏结束*/ EndGame(); } /**********************************************************/
/*界面初始化，数据初始化*/ void Init(void) { int i,j; char *Msg[]= { "Player1 key:", " UP----w", " DOWN--s", " LEFT--a", " RIGHT-d", " DO----space", "", "Player2 key:", " UP----up", " DOWN--down", " LEFT--left", " RIGHT-right", " DO----ENTER", "", "exit game:", " ESC", NULL, };
/*先手方为1号玩家*/ gPlayOrder = CHESS1; /*棋盘数据清零, 即棋盘上各点开始的时候都没有棋子*/ for(i=0;i19;i++) for(j=0;j19;j++) gChessBoard[i][j]=CHESSNULL; /*光标初始位置*/ gCursor.x=gCursor.y=0;
/*画棋盘*/ textmode(C40); DrawMap();
/*显示操作键说明*/ i=0; textcolor(BROWN); while(Msg[i]!=NULL) { gotoxy(25,3+i); cputs(Msg[i]); i++; }
/*显示当前行棋方*/ ShowOrderMsg(gPlayOrder); /*光标移至棋盘的左上角点处*/ gotoxy(gCursor.x+MAPXOFT,gCursor.y+MAPYOFT); }
/*画棋盘*/ void DrawMap(void) { int i,j;
clrscr();
for(i=0;i19;i++) for(j=0;j19;j++) DrawCross(i,j);
}
/*画棋盘上的交叉点*/ void DrawCross(int x,int y) { gotoxy(x+MAPXOFT,y+MAPYOFT); /*交叉点上是一号玩家的棋子*/ if(gChessBoard[x][y]==CHESS1) { textcolor(LIGHTBLUE); putch(CHESS1); return; } /*交叉点上是二号玩家的棋子*/ if(gChessBoard[x][y]==CHESS2) { textcolor(LIGHTBLUE); putch(CHESS2); return; }
textcolor(GREEN);
/*左上角交叉点*/ if(x==0y==0) { putch(CROSSLU); return; }
/*左下角交叉点*/ if(x==0y==18) { putch(CROSSLD); return; } /*右上角交叉点*/ if(x==18y==0) { putch(CROSSRU); return; }
/*右下角交叉点*/ if(x==18y==18) { putch(CROSSRD); return; }
/*左边界交叉点*/ if(x==0) { putch(CROSSL); return; }
/*右边界交叉点*/ if(x==18) { putch(CROSSR); return; }
/*上边界交叉点*/ if(y==0) { putch(CROSSU); return; }
/*下边界交叉点*/ if(y==18) { putch(CROSSD); return; }
/*棋盘中间的交叉点*/ putch(CROSS); }
/*交换行棋方*/ int ChangeOrder(void) { if(gPlayOrder==CHESS1) gPlayOrder=CHESS2; else gPlayOrder=CHESS1;
return(gPlayOrder); }
/*获取按键值*/ int GetKey(void) { char lowbyte; int press;
while (bioskey(1) == 0) ;/*如果用户没有按键，空循环*/
press=bioskey(0); lowbyte=press0xff; press=press0xff00 + toupper(lowbyte); return(press); }
/*落子错误处理*/ void DoError(void) { sound(1200); delay(50); nosound(); }
/*赢棋处理*/ void DoWin(int Order) { sound(1500);delay(100); sound(0); delay(50); sound(800); delay(100); sound(0); delay(50); sound(1500);delay(100); sound(0); delay(50); sound(800); delay(100); sound(0); delay(50); nosound();
textcolor(RED+BLINK); gotoxy(25,20); if(Order==CHESS1) cputs("PLAYER1 WIN!"); else cputs("PLAYER2 WIN!"); gotoxy(25,21); cputs(" /"\\^+^/"); getch(); }
/*走棋*/ int ChessGo(int Order,struct point Cursor) { /*判断交叉点上有无棋子*/ if(gChessBoard[Cursor.x][Cursor.y]==CHESSNULL) { /*若没有棋子, 则可以落子*/ gotoxy(Cursor.x+MAPXOFT,Cursor.y+MAPYOFT); textcolor(LIGHTBLUE); putch(Order); gotoxy(Cursor.x+MAPXOFT,Cursor.y+MAPYOFT); gChessBoard[Cursor.x][Cursor.y]=Order; return TRUE; } else return FALSE; }
/*判断当前行棋方落子后是否赢棋*/ int JudgeWin(int Order,struct point Cursor) { int i; for(i=0;i4;i++) /*判断在指定方向上是否有连续5个行棋方的棋子*/ if(JudgeWinLine(Order,Cursor,i)) return TRUE; return FALSE; }
/*判断在指定方向上是否有连续5个行棋方的棋子*/ int JudgeWinLine(int Order,struct point Cursor,int direction) { int i; struct point pos,dpos; const int testnum = 5; int count;
switch(direction) { case 0:/*在水平方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y; dpos.x=1; dpos.y=0; break; case 1:/*在垂直方向*/ pos.x=Cursor.x; pos.y=Cursor.y-(testnum-1); dpos.x=0; dpos.y=1; break; case 2:/*在左下至右上的斜方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y+(testnum-1); dpos.x=1; dpos.y=-1; break; case 3:/*在左上至右下的斜方向*/ pos.x=Cursor.x-(testnum-1); pos.y=Cursor.y-(testnum-1); dpos.x=1; dpos.y=1; break; }
count=0; for(i=0;itestnum*2+1;i++) { if(pos.x=0pos.x=18pos.y=0pos.y=18) { if(gChessBoard[pos.x][pos.y]==Order) { count++; if(count=testnum) return TRUE; } else count=0; } pos.x+=dpos.x; pos.y+=dpos.y; }
return FALSE; }
/*移动光标*/ void MoveCursor(int Order,int press) { switch(press) { case PLAY1UP: if(Order==CHESS1gCursor.y0) gCursor.y--; break; case PLAY1DOWN: if(Order==CHESS1gCursor.y18) gCursor.y++; break; case PLAY1LEFT: if(Order==CHESS1gCursor.x0) gCursor.x--; break; case PLAY1RIGHT: if(Order==CHESS1gCursor.x18) gCursor.x++; break;
case PLAY2UP: if(Order==CHESS2gCursor.y0) gCursor.y--; break; case PLAY2DOWN: if(Order==CHESS2gCursor.y18) gCursor.y++; break; case PLAY2LEFT: if(Order==CHESS2gCursor.x0) gCursor.x--; break; case PLAY2RIGHT: if(Order==CHESS2gCursor.x18) gCursor.x++; break; }
gotoxy(gCursor.x+MAPXOFT,gCursor.y+MAPYOFT); }
/*游戏结束处理*/ void EndGame(void) { textmode(C80); }
/*显示当前行棋方*/ void ShowOrderMsg(int Order) { gotoxy(6,MAPYOFT+20); textcolor(LIGHTRED); if(Order==CHESS1) cputs("Player1 go!"); else cputs("Player2 go!");
gotoxy(gCursor.x+MAPXOFT,gCursor.y+MAPYOFT); }
/*落子正确处理*/ void DoOK(void) { sound(500); delay(70); sound(600); delay(50); sound(1000); delay(100); nosound(); }
/*检查用户的按键类别*/ int CheckKey(int press) { if(press==ESCAPE) return KEYEXIT;/*是退出键*/
else if ( ( press==PLAY1DO gPlayOrder==CHESS1) || ( press==PLAY2DO gPlayOrder==CHESS2) ) return KEYFALLCHESS;/*是落子键*/
else if ( press==PLAY1UP || press==PLAY1DOWN || press==PLAY1LEFT || press==PLAY1RIGHT || press==PLAY2UP || press==PLAY2DOWN || press==PLAY2LEFT || press==PLAY2RIGHT ) return KEYMOVECURSOR;/*是光标移动键*/
else return KEYINVALID;/*按键无效*/ }
Q2: C语言扫雷点到空白地方一次打开一大片的算法求解
能把代码再多给点吗？
那8个递归其实可以用两个for来做的
void ClickBlank(int x, int y) {
int i, j;
if (isBlank(x, y)) {
ShowAroundBlock(x, y);
for (i = -1; i = 1; i++) {
for (j = -1; j = 1; j++) {
if (i == 0 j == 0) continue;
ClickBlank(x + i, y + j);
}
}
}
}
而且你应该设访问标记，已经ClickBlank的格子不应再调用ClickBlank
Q3: 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();
}
}
Q4: 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。
Q5: 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;
}
