使用C语言如何实现一个俄罗斯方块小游戏-创新互联
使用C语言如何实现一个俄罗斯方块小游戏？很多新手对此不是很清楚，为了帮助大家解决这个难题，下面小编将为大家详细讲解，有这方面需求的人可以来学习下，希望你能有所收获。
具体内容如下
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#define TTY_PATH "/dev/tty"
#define STTY_ON "stty raw -echo -F"
#define STTY_OFF "stty -raw echo -F"
int map[21][14];
char direct;
int node[7][4][16]={
{{0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0},//长方形
{0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0},
{0,0,0,0,1,1,1,1,0,0,0,0,0,0,0,0},
{0,1,0,0,0,1,0,0,0,1,0,0,0,1,0,0}},
{{1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0},//正方形
{1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0},
{1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0},
{1,1,0,0,1,1,0,0,0,0,0,0,0,0,0,0}},
{{0,1,0,0,1,1,1,0,0,0,0,0,0,0,0,0},//3边加一中点
{0,1,0,0,0,1,1,0,0,1,0,0,0,0,0,0},
{0,0,0,0,1,1,1,0,0,1,0,0,0,0,0,0},
{0,1,0,0,1,1,0,0,0,1,0,0,0,0,0,0}},
{{0,1,1,0,0,1,0,0,0,1,0,0,0,0,0,0},//右锄头型
{0,0,0,0,1,1,1,0,0,0,1,0,0,0,0,0},
{0,1,0,0,0,1,0,0,1,1,0,0,0,0,0,0},
{1,0,0,0,1,1,1,0,0,0,0,0,0,0,0,0}},
{{1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0},//左锄头型
{0,0,1,0,1,1,1,0,0,0,0,0,0,0,0,0},
{0,1,0,0,0,1,0,0,0,1,1,0,0,0,0,0},
{0,0,0,0,1,1,1,0,1,0,0,0,0,0,0,0}},
{{0,1,0,0,0,1,1,0,0,0,1,0,0,0,0,0},//右曲折型
{0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0},
{0,1,0,0,0,1,1,0,0,0,1,0,0,0,0,0},
{0,1,1,0,1,1,0,0,0,0,0,0,0,0,0,0}},
{{0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0},//左曲折型
{1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0},
{0,1,0,0,1,1,0,0,1,0,0,0,0,0,0,0},
{1,1,0,0,0,1,1,0,0,0,0,0,0,0,0,0}}
};
typedef struct block
{
int x;
int y;
int blockType;
int blockDirect;
}Block;
Block bl;
void init_map()//初始化边框
{
int i,j;
for(i=0; i<21; i++)
for(j=0; j<14; j++)
{
if(j==0 || j==13)
map[i][j] = 200;
else if(i==20)
map[i][j] = 201;
else
map[i][j] = 0;
}
}
void new_block()//生成随机的俄罗斯方块
{
int blockType = rand()%7;
int blockDirect = rand()%4;
int x = 1;
int y = 5;
bl.x = x;
bl.y = y;
bl.blockType = blockType;
bl.blockDirect = blockDirect;
}
void input()//将移动后的俄罗斯方块，导入地图中作标记
{
int i, j;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j]==1)
{
map[bl.x+i][bl.y+j] = 1;
}
}
void output()//移动时，将之前俄罗斯方块在地图信息清空。
{
int i, j;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j]==1)
{
map[bl.x+i][bl.y+j] = 0;
}
}
void change()//俄罗斯方格在碰撞后融入，固定
{
int i, j;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j]==1)
{
map[bl.x+i][bl.y+j] = 10;
}
for(j=1; j<13; j++)
if(map[5][j] == 10)
{
system("clear");
printf("game over !!!!!!!!!\n");
exit(1);
}
}
void print_map()//打印地图，显示信息
{
int i,j;
for(i=5; i<21; i++)
{
for(j=0; j<14; j++)
{
if(map[i][j]==200)//左右边界
printf("#");
else if(map[i][j]==201)//下边界
printf(" # ");
else if(map[i][j]==0)//空白地
printf(" ");
else if(map[i][j]==1)//移动的俄罗斯方块
printf(" * ");
else if(map[i][j]==10)//固定的俄罗斯方块
printf(" @ ");
}
printf("\n");
}
}
void delLine(int n)//消行
{
int i,j;
for(j = 1; j<13; j++)
map[n][j] = 0;
for(i = n; i>5 ; i--)
for(j = 1; j<13; j++)
if(map[i-1][j] != 1)
map[i][j] = map[i-1][j];
}
void isFillLine()//是否满足消行条件
{
int i,j;
int fals;
for(i=19; i>5; i--)
{
fals = 1;
for(j=1; j<13; j++)
{
if(map[i][j] != 10)
{
fals = 0;
continue;
}
}
if(fals)
{
delLine(i);
}
}
}
void down()//下移
{
int i, j;
int fale = 1;
for(i=3; i>=0; i--)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j] == 1)
if(map[bl.x+i+1][bl.y+j] == 10 || map[bl.x+i+1][bl.y+j] == 201)
{
change();
fale = 0;
new_block();
isFillLine();
return;
}
if(fale)
{
output();
bl.x += 1;
input();
}
}
void right()//右移
{
int i, j;
int fale = 1;
for(i=3; i>=0; i--)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j] == 1)
if(map[bl.x+i][bl.y+j+1] == 10 || map[bl.x+i][bl.y+j+1] == 200)
{
fale = 0;
return;
}
if(fale)
{
output();
bl.y += 1;
input();
}
}
void left()//左移
{
int i, j;
int fale = 1;
for(i=3; i>=0; i--)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j] == 1)
if(map[bl.x+i][bl.y+j-1] == 10 || map[bl.x+i][bl.y+j-1] == 200)
{
fale = 0;
return;
}
if(fale)
{
output();
bl.y -= 1;
input();
}
}
void change_block()//俄罗斯方块变形
{
int i,j;
output();
int fals = 1;
bl.blockDirect += 1;
bl.blockDirect %= 4;
for(i=0; i<4; i++)
for(j=0; j<4; j++)
if(node[bl.blockType][bl.blockDirect][i*4+j]==1)
if(map[bl.x+i][bl.y+j] != 0 )
{
fals = 0;
break;
}
if(fals)
{
input();
}else
{
bl.blockDirect -= 1;
input();
}
}
char in_direct()//非堵塞输入
{
fd_set fd;
struct timeval tv;
char ch;
FD_ZERO(&fd);
FD_SET(0, &fd);
tv.tv_sec = 0;
tv.tv_usec = 10;
if(select(1, &fd ,NULL, NULL, &tv) > 0)
{
ch = getchar();
}
return ch;
}
int main()//q 退出游戏，a,d 左右移动，空格变形
{
srand(time(NULL));
init_map();
new_block();
input();
char ch;
int num = 0;
while(1)
{
usleep(500000);
system(STTY_ON TTY_PATH);
ch = in_direct();
system(STTY_OFF TTY_PATH);
system("clear");
if(ch == 'a' && num <= 1)
{
left();
print_map();
num++;
continue;
}else if(ch == 'd' && num <= 1)
{
right();
print_map();
num++;
continue;
}else if(ch == ' ' && num <= 1 )
{
change_block();
print_map();
num++;
continue;
}else if(ch == 'q')
{
system("clear");
printf("gave over!!!!!\n");
exit(0);
}
down();
print_map();
num = 0;
}
return 0;
}看完上述内容是否对您有帮助呢？如果还想对相关知识有进一步的了解或阅读更多相关文章，请关注创新互联网站建设公司行业资讯频道，感谢您对创新互联建站的支持。
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