请帮助贪吃蛇决定如何移动才能重获自由！！！求指点

慢慢看吧然后修改程序。 #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 0x011bint i,key;int score=0;/*得分*/int gamespeed=50000;/*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;elseif(key==UP&&snake.direction!=4)/*判断是否往相反的方向移动*/snake.direction=3;elseif(key==RIGHT&&snake.direction!=2)snake.direction=1;elseif(key==LEFT&&snake.direction!=1)snake.direction=2;elseif(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();}

请帮助贪吃蛇决定如何移动才能重获自由！！！求指点

2回答