掃雷編程c語言,c語言編寫掃雷

本文目錄一覽：

• 1、掃雷C語言
• 2、如何用C語言編程 掃雷！~
• 3、C語言掃雷遊戲源代碼

掃雷C語言

#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(“輸入個數：\n”);

scanf(“%d”,num);

printf(“輸入數據(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);

}

演算法思想：

1、如果有輸入數字是3則輸出數字中對應上中下都必為1

2、輸出數組中只有為0的才能為1；

3、用回溯法判斷成立條件，成功則輸出。

如何用C語言編程 掃雷！~

俄羅斯方快

掃雷

#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();

}

}

C語言掃雷遊戲源代碼

“掃雷”小遊戲C代碼

#includestdio.h

#includemath.h

#includetime.h

#includestdlib.h

main( )

{char a[102][102],b[102][102],c[102][102],w;

int i,j;  /*循環變數*/

int x,y,z[999];  /*雷的位置*/

int t,s;  /*標記*/

int m,n,lei;  /*計數*/

int u,v;  /*輸入*/

int hang,lie,ge,mo;  /*自定義變數*/

srand((int)time(NULL));  /*啟動隨機數發生器*/

leb1:  /*選擇模式*/

printf(“\n   請選擇模式：\n   1.標準  2.自定義\n”);

scanf(“%d”,mo);

if(mo==2)  /*若選擇自定義模式，要輸入三個參數*/

{do

{t=0; printf(“請輸入\n行數 列數 雷的個數\n”);

scanf(“%d%d%d”,hang,lie,ge);

if(hang2){printf(“行數太少\n”); t=1;}

if(hang100){printf(“行數太多\n”);t=1;}

if(lie2){printf(“列數太少\n”);t=1;}

if(lie100){printf(“列數太多\n”);t=1;}

if(ge1){printf(“至少要有一個雷\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;}  /*否則就是選擇了標準模式(默認參數)*/

for(i=1;i=ge;i=i+1)  /*確定雷的位置*/

{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)  /*計算b中數字*/

{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;  /*以下是遊戲設計*/

do

{leb2:  /*輸出*/

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(” 剩餘雷個數”);

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);  /*輸入*/

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:  /*打開0區*/

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)  /*檢測0區*/

{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;  /*檢查結束*/

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)  /*遊戲結束*/

{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(” 剩餘雷個數”);

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(”    遊戲結束!\n”);

printf(”    重玩請輸入1\n”);

t=0;

scanf(“%d”,t);

if(t==1)goto leb1;

}

/*註：在DEV c++上運行通過。行號和列號都從0開始，比如要確定第0行第9列不是「雷」，就在0和9中間加入一個字母，可以輸入【0a9】三個字元再按回車鍵。3行7列不是雷，則輸入【3a7】回車；第8行第5列是雷，就輸入【8#5】回車，9行0列是雷則輸入【9#0】並回車*/