求，贪吃蛇 C语言代码 及其每一步的讲解

Answer 1

/* 贪吃蛇程序 by champking */

#define N 200

#include <graphics.h>
#include <stdlib.h>
#include <dos.h>

#define LEFT 0x4b00
#define RIGHT 0x4d00
#define DOWN 0x5000
#define UP 0x4800
#define ESC 0x011b

int i,key;
int score = 0;/*得分*/
int gamespeed = 100000;/*游戏速度自己调整*/

struct Food
{
int x;/*食物的横坐标*/
int y;/*食物的纵坐标*/
int yes;/*判断是否要出现食物的变量*/
}food;/*食物的结构体*/

struct Snake
{
int x[N];
int y[N];
int node;/*蛇的节数*/
int direction;/*蛇移动方向*/
int life;/* 蛇的生命,0活着,1死亡*/
}snake;

void Init(void);/*图形驱动*/
void Close(void);/*图形结束*/
void DrawK(void);/*开始画面*/
void GameOver(void);/*结束游戏*/
void GamePlay(void);/*玩游戏具体过程*/
void PrScore(void);/*输出成绩*/

/*主函数*/
void main(void)
{
Init();/*图形驱动*/
DrawK();/*开始画面*/
GamePlay();/*玩游戏具体过程*/
Close();/*图形结束*/
}

/*图形驱动*/
void Init(void)
{
int gd = DETECT, gm;
initgraph(&gd, &gm, "c:\\tc");
cleardevice();
}
/*开始画面,左上角坐标为(50,40),右下角坐标为(610,460)的围墙*/
void DrawK(void)
{
/*setbkcolor(LIGHTGREEN);*/
setcolor(11);
setlinestyle(SOLID_LINE, 0, THICK_WIDTH);/*设置线型*/

for(i = 50; i <= 600; i += 10)/*画围墙*/
{
rectangle(i, 40, i + 10, 49); /*上边*/
rectangle(i, 451, i + 10, 460);/*下边*/
}

for(i = 40; i <= 450; i += 10)
{
rectangle(50, i, 59, i + 10); /*左边*/
rectangle(601, i, 610, i + 10);/*右边*/
}
}
/*玩游戏具体过程*/
void GamePlay(void)
{
randomize();/*随机数发生器*/
food.yes = 1;/*1表示需要出现新食物,0表示已经存在食物*/
snake.life = 0;/*活着*/
snake.direction = 1;/*方向往右*/
snake.x[0] = 100; snake.y[0] = 100;/*蛇头*/
snake.x[1] = 110; snake.y[1] = 100;
snake.node = 2;/*节数*/
PrScore();/*输出得分*/

while(1)/*可以重复玩游戏,压ESC键结束*/
{
while(!kbhit())/*在没有按键的情况下,蛇自己移动身体*/
{
if(food.yes == 1)/*需要出现新食物*/
{
food.x = rand() % 400 + 60;
food.y = rand() % 350 + 60;

while(food.x % 10 != 0)/*食物随机出现后必须让食物能够在整格内,这样才可以让蛇吃到*/
food.x++;
while(food.y % 10 != 0)
food.y++;
food.yes = 0;/*画面上有食物了*/
}

if(food.yes == 0)/*画面上有食物了就要显示*/
{
setcolor(GREEN);
rectangle(food.x, food.y, food.x + 10, food.y - 10);
}

for(i = snake.node - 1; i > 0; i--)/*蛇的每个环节往前移动,也就是贪吃蛇的关键算法*/
{
snake.x[i] = snake.x[i-1];
snake.y[i] = snake.y[i-1];
}

/*1,2,3,4表示右,左,上,下四个方向,通过这个判断来移动蛇头*/
switch(snake.direction)
{
case 1: snake.x[0] += 10; break;
case 2: snake.x[0] -= 10; break;
case 3: snake.y[0] -= 10; break;
case 4: snake.y[0] += 10; break;
}

for(i = 3; i < snake.node; i++)/*从蛇的第四节开始判断是否撞到自己了,因为蛇头为两节,第三节不可能拐过来*/
{
if(snake.x[i] == snake.x[0] && snake.y[i] == snake.y[0])
{
GameOver();/*显示失败*/
snake.life = 1;
break;
}
}

if(snake.x[0]<55||snake.x[0]>595||snake.y[0]<55||
snake.y[0]>455)/*蛇是否撞到墙壁*/
{
GameOver();/*本次游戏结束*/
snake.life=1; /*蛇死*/
}

if(snake.life == 1)/*以上两种判断以后,如果蛇死就跳出内循环,重新开始*/
break;

if(snake.x[0] == food.x && snake.y[0] == food.y)/*吃到食物以后*/
{
setcolor(0);/*把画面上的食物东西去掉*/
rectangle(food.x, food.y, food.x + 10, food.y - 10);
snake.x[snake.node] =- 20; snake.y[snake.node] =- 20;
/*新的一节先放在看不见的位置,下次循环就取前一节的位置*/
snake.node++;/*蛇的身体长一节*/
food.yes = 1;/*画面上需要出现新的食物*/
score += 10;
PrScore();/*输出新得分*/
}

setcolor(4);/*画出蛇*/

for(i = 0; i < snake.node; i++)
rectangle(snake.x[i], snake.y[i], snake.x[i] + 10,
snake.y[i] - 10);

delay(gamespeed);

setcolor(0);/*用黑色去除蛇的的最后一节*/
rectangle(snake.x[snake.node-1], snake.y[snake.node-1],
snake.x[snake.node-1] + 10, snake.y[snake.node - 1] - 10);
} /*endwhile(！kbhit)*/

if(snake.life == 1)/*如果蛇死就跳出循环*/
break;

key = bioskey(0);/*接收按键*/

if(key == ESC)/*按ESC键退出*/
break;
else
if(key == UP&&snake.direction!=4)
/*判断是否往相反的方向移动*/
snake.direction=3;
else
if(key == RIGHT &&snake.direction != 2)
snake.direction=1;
else
if(key == LEFT && snake.direction != 1)
snake.direction = 2;
else
if(key == DOWN && snake.direction != 3)
snake.direction = 4;
}/*endwhile(1)*/
}

/*游戏结束*/
void GameOver(void)
{
cleardevice();
PrScore();
setcolor(RED);
settextstyle(0, 0, 4);
outtextxy(200, 200, "GAME OVER");
getch();
}

/*输出成绩*/
void PrScore(void)
{
char str[10];
setfillstyle(SOLID_FILL, YELLOW);
bar(50, 15, 220, 35);
setcolor(6);
settextstyle(0,0,2);
sprintf(str, "score:%d", score);
outtextxy(55, 20, str);
}

/*图形结束*/
void Close(void)
{
getch();
closegraph();
}

参考资料：http://zhidao.baidu.com/question/107200932.html

Answer 2

楼上正解，比较详细。