学学写个智能蛇

引言

我先放一张传说中的贪吃智能蛇

言归正传

首先，我写了个能够每秒自动走的蛇作为原型

#include<stdio.h>#include<conio.h>#include<string.h>#include<stdlib.h>#include<time.h>#define FOR(i,j,k) for(i=j;i<=k;++i)int d[1000][1000];char map[1000][1000] = { ' ' };int fx, fy, lx, ly;int len = 5;int diffcuilty;void pm();int n, m, w, h;//W 1 ;A 2 ;S 3 ;D 4void CreateTheBackground(int x, int y){    int i, j, k;    FOR(i, 1, w)        FOR(j, 1, h)    {        map[i][j] = ' '; d[i][j] = 0;    }    FOR(i, 1, 5) {        map[1][i] = '*';        d[1][i] = 1;    }    fx = 1; fy = 5;    map[1][5] = '%';    lx = ly = 1;    FOR(i, 0, h + 1) { map[0][i] = '#'; map[w + 1][i] = '#'; }    FOR(i, 0, w + 1) { map[i][0] = '#'; map[i][h + 1] = '#'; }    pm();}void pm(){    srand(time(NULL));    int i, j;    i = rand() % w + 1;    j = rand() % h + 1;    while (map[i][j] != ' ')    {        i = rand() % w + 1;        j = rand() % h + 1;    }    map[i][j] = '$';}int getinp()//W 1 ;A 2 ;S 3 ;D 4{    int dr = 0;    char c = 0;    int tle = 1;    int start = clock();    while (clock() - start <= diffcuilty * 50)    {        if (kbhit()) c = getch();        if (c != 0) { tle--; }        if (c == 'w') dr = 1;        if (c == 'a') dr = 2;        if (c == 's') dr = 3;        if (c == 'd') dr = 4;        if (clock() - start > diffcuilty * 50) return dr;    }    if (c == 'w') dr = 1;    if (c == 'a') dr = 2;    if (c == 's') dr = 3;    if (c == 'd') dr = 4;    return dr;}void pp(){    int i, j;    system("cls");    for (i = h + 1; i >= 0; --i)    {        FOR(j, 0, w + 1)            printf("%c", map[j][i]);        printf("\n");    }}void dead(){    int t;    printf("Failture with the score:%d\n", len);    printf("\n\n\nPress any key to exit.");    getchar();    exit(0);}void movement(int dr){    int tx, ty, t;    d[fx][fy] = dr;    int bj = 0;    map[fx][fy] = '*';    if (dr == 1) fy++;    if (dr == 2) fx--;    if (dr == 3) fy--;    if (dr == 4) fx++;    if (map[fx][fy] == '$') { bj = 1; }    if (map[fx][fy] == '*' || map[fx][fy] == '#') { dead(); return; }    map[fx][fy] = '%';    if (bj) {        pm();        len++;        return;    }    t = d[lx][ly];    map[lx][ly] = ' ';    if (t == 1) ly++;    if (t == 2) lx--;    if (t == 3) ly--;    if (t == 4) lx++;}int u = 1;void done()//W 1 ;A 2 ;S 3 ;D 4{    int d = getinp();    if (d != 0) u = d;    movement(u);    pp();}int main(){BACK:    printf("Firstly, please input the width and height of the area:\n(The width and height are recommanded more than 5)\n");ERROR:    scanf("%d%d", &w, &h);    if (w <= 5 || h <= 5) {        printf("Input illegal,please input again\n");        goto ERROR;    }ERROR2:    printf("Please choose a diffcuilty of this game:\nThe diffcuilty from easy to diffcuilt show 1~16\n");    scanf("%d", &diffcuilty);    if (!(diffcuilty >= 1 && diffcuilty <= 16)) {        printf("Input illegal,please input again\n");        goto ERROR2;    }    diffcuilty = 17 - diffcuilty;    printf("\n");    system("cls");    CreateTheBackground(w, h);    pp();    system("pause");    int d = 1;    while (1)    {        done();    }    return 0;}

然后写了第一次作业

#include<stdio.h>#include<time.h>#define FOR(i,j,k) for(i=j;i<=k;++i)#define SNAKE_MAX_LENGTH 20#define SNAKE_HEAD 'H'#define SNAKE_BODY 'X'#define BLANK_CELL ' '#define SNAKE_FOOD '$'#define WALL_CELL '*'int life = 1;int snakey[100] = { 5,4,3,2,1 };int snakex[100] = { 1,1,1,1,1 };int d[1000][1000];char map[12][12] = { "************",//da biao"*XXXXH     *","* ******** *","*          *","*   ****   *","*   ****   *","*   ****   *","*          *","* ******** *","*          *","************" };int fx, fy, lx, ly;int len = 5;int diffcuilty;void pm();int n, m, w, h;//W 1 ;A 2 ;S 3 ;D 4void CreateTheBackground(){    int i, j, k;    w = 10; h = 10;    FOR(i, 1, w)        FOR(j, 1, h)    {        if (map[i][j] != '*') {            map[i][j] = ' ';            d[i][j] = 0;        }    }    FOR(i, 1, 4) {        map[i][10] = 'X';        d[i][10] = 4;    }    fx = 5; fy = 10;    d[5][10] = 4;    map[5][10] = 'H';    lx = 1; ly = 10;    FOR(i, 0, h + 1) { map[0][i] = '*'; map[w + 1][i] = '*'; }    FOR(i, 0, w + 1) { map[i][0] = '*'; map[i][h + 1] = '*'; }    pm();}void pm(){    srand(time(NULL));    int i, j;    i = rand() % w + 1;    j = rand() % h + 1;    while (map[i][j] != ' ')    {        i = rand() % w + 1;        j = rand() % h + 1;    }    map[i][j] = '$';}int getinp()//W 1 ;A 2 ;S 3 ;D 4{    char c;    int dr = 0;    c = getch();    if (c == 'W') dr = 1;    if (c == 'A') dr = 2;    if (c == 'S') dr = 3;    if (c == 'D') dr = 4;    return dr;}void pp(){    int i, j;    for (i = h + 1; i >= 0; --i)    {        FOR(j, 0, w + 1)            printf("%c", map[j][i]);        printf("\n");    }}void dead(){    int t;    printf("Game Over!!!");    //getchar();}void movement(int dr){    int tx, ty, t;    d[fx][fy] = dr;    int bj = 0;    map[fx][fy] = 'X';    if (dr == 1) fy++;    if (dr == 2) fx--;    if (dr == 3) fy--;    if (dr == 4) fx++;    if (map[fx][fy] == '$') { bj = 1; }    if (map[fx][fy] == 'X' || map[fx][fy] == '*') { life = 0; return; }    map[fx][fy] = 'H';    if (bj) {        pm();//fang zhi jian qian        len++;        return;    }    t = d[lx][ly];    map[lx][ly] = ' ';    if (t == 1) ly++;    if (t == 2) lx--;    if (t == 3) ly--;    if (t == 4) lx++;}int u = 1;void done()//W 1 ;A 2 ;S 3 ;D 4{    int d = getinp();    if (d != 0) u = d;    movement(u);    pp();}int main(){    CreateTheBackground();//chuang jian bei jing     pp();//shu chu    int ch;    while (life)    {        ch = getinp();//shu ru        switch (ch)//gen ju shu ru de fang xiang yi dong        {        case 1:movement(1); break;        case 2:movement(2); break;        case 3:movement(3); break;        case 4:movement(4); break;        }        if (life) pp();    }    dead();//shu chu si wang     return 0;}

智能蛇的实现其实只需要写出一个函数替代getinp（）函数并每秒执行一次即可

下面着重讲下搜索策略
方案1.0
在一个矩形中，每一时刻有一个食物，贪吃蛇要在不撞到自己的条件下，
找到一条路(未必要最优)，然后沿着这条路运行，去享用它的美食

显然这个仅仅一个bfs即可实现，此蛇很容易被自己直接绕住

方法2.0
一开始，蛇很短(初始化长度为1)，它看到了一个食物， 使用BFS得到矩形中每个位置到达食物的最短路径长度。在没有蛇身阻挡下， 就是曼哈顿距离。然后，我要先判断一下，贪吃蛇这一去是否安全。 所以我需要一条虚拟的蛇，它每次负责去探路。如果安全，才让真正的蛇去跑。 当然，虚拟的蛇是不会绘制出来的，它只负责模拟探路。那么， 怎么定义一个布局是安全的呢？ 如果你把文章开头那张动态图片中蛇的销魂走位好好的看一下， 会发现即使到最后蛇身已经很长了，它仍然没事一般地走出了一条路。而且， 是跟着蛇尾走的！嗯，这个其实不难解释，蛇在运动的过程中，消耗蛇身， 蛇尾后面总是不断地出现新的空间。蛇短的时候还无所谓，当蛇一长， 就会发现，要想活下来，基本就只能追着蛇尾跑了。在追着蛇尾跑的过程中， 再去考虑能否安全地吃到食物。

本人写的代码
基本实现了上述的算法

#include<stdio.h>#include<time.h>#define FOR(i,j,k) for(i=j;i<=k;++i)#define SNAKE_MAX_LENGTH 20#define SNAKE_HEAD 'H'#define SNAKE_BODY 'X'#define BLANK_CELL ' '#define SNAKE_FOOD '$'#define WALL_CELL '*'#define mal 0x7fffffffint life = 1;int snakey[100] = { 5,4,3,2,1 };int snakex[100] = { 1,1,1,1,1 };int d[1000][1000];char map[12][12] = { "************",//da biao"*XXXXH     *","*          *","*          *","*          *","*          *","*          *","*          *","*          *","*          *","************" };int fx, fy, lx, ly;int len = 5;int mx,my;int diffcuilty;void pm();int n, m, w, h;//W 1 ;A 2 ;S 3 ;D 4char cp[12][12];int f[12][12];int qx[1000],qy[1000];int qboo[12][12];int dxx[5]={0,0,-1,0,1};int dyy[5]={0,1,0,-1,0};void bfs(int lx,int ly)//yi (lx,ly) kuo zhang chu bfs ju li (man ha dun ju li){    int ux=lx,uy=ly;int tx,ty;    int i,j,k;    int be=0,en=0;    FOR(i,1,10)        FOR(j,1,10)            {                qboo[i][j]=1;                f[i][j]=mal;            }    qx[0]=lx;qy[0]=ly;qboo[lx][ly]=0;    f[lx][ly]=0;    while(be<=en)    {        k=be;be++;        ux=qx[k];        uy=qy[k];        qboo[ux][uy]=0;        FOR(i,1,4)        {            tx=dxx[i]+ux;            ty=dyy[i]+uy;            if ((cp[tx][ty]==' '||cp[tx][ty]=='H')&&qboo[tx][ty])                {                    f[tx][ty]=f[ux][uy]+1;                    en++;                    qx[en]=tx;                    qy[en]=ty;                }        }    }}void cop()//jiang mu qian zhuang tai fu zhi gei cp[][];{    int i,j;    FOR(i,0,11)        FOR(j,0,11)            cp[i][j]=map[i][j];}int ddd()//qiu fang xiang jie{    cop();int i,j,k=10000;int l;    bfs(mx,my);    if (f[fx][fy]!=mal)//ruo she tou ke yi di da        {            FOR(i,1,4)                if (f[fx+dxx[i]][fy+dyy[i]]<k)                    {                        k=f[fx+dxx[i]][fy+dyy[i]];                        j=i;                    }            return j;        }    else    {        bfs(lx,ly);        k=-1;        FOR(i,1,4)                if (f[fx+dxx[i]][fy+dyy[i]]>k&&f[fx+dxx[i]][fy+dyy[i]]!=mal)                    {                        k=f[fx+dxx[i]][fy+dyy[i]];                        j=i;                    }            return j;    }}void CreateTheBackground(){    int i, j, k;    w = 10; h = 10;    FOR(i, 1, w)        FOR(j, 1, h)    {        if (map[i][j] != '*') {            map[i][j] = ' ';            d[i][j] = 0;        }    }    FOR(i, 1, 4) {        map[i][10] = 'X';        d[i][10] = 4;    }    fx = 5; fy = 10;    d[5][10] = 4;    map[5][10] = 'H';    lx = 1; ly = 10;    FOR(i, 0, h + 1) { map[0][i] = '*'; map[w + 1][i] = '*'; }    FOR(i, 0, w + 1) { map[i][0] = '*'; map[i][h + 1] = '*'; }    pm();}void pm(){    srand(time(NULL));    int i, j;    i = rand() % w + 1;    j = rand() % h + 1;    while (map[i][j] != ' ')    {        i = rand() % w + 1;        j = rand() % h + 1;    }    mx=i;    my=j;    map[i][j] = '$';}int getinp()//W 1 ;A 2 ;S 3 ;D 4{    int start = clock();    int dr;    while (clock() - start <= 1000);    dr=ddd();    return dr;}void pp(){    int i, j;    system("cls");    for (i = h + 1; i >= 0; --i)    {        FOR(j, 0, w + 1)            printf("%c", map[j][i]);        printf("\n");    }}void dead(){    int t;    printf("Game Over!!!");    //getchar();}void movement(int dr){    int tx, ty, t;    d[fx][fy] = dr;    int bj = 0;    map[fx][fy] = 'X';    if (dr == 1) fy++;    if (dr == 2) fx--;    if (dr == 3) fy--;    if (dr == 4) fx++;    if (map[fx][fy] == '$') { bj = 1; }    if (map[fx][fy] == 'X' || map[fx][fy] == '*') { life = 0; return; }    map[fx][fy] = 'H';    if (bj) {        pm();//fang zhi jian qian        len++;        return;    }    t = d[lx][ly];    map[lx][ly] = ' ';    if (t == 1) ly++;    if (t == 2) lx--;    if (t == 3) ly--;    if (t == 4) lx++;}int u = 1;void done()//W 1 ;A 2 ;S 3 ;D 4{    int d = getinp();    if (d != 0) u = d;    movement(u);    pp();}int main(){    CreateTheBackground();//chuang jian bei jing     pp();//shu chu    int ch;    int las=4;    while (life)    {        ch = getinp();//shu ru        if (!(ch>=1&&ch<=4) ) ch=las;        las=ch;        switch (ch)//gen ju shu ru de fang xiang yi dong        {        case 1:movement(1); break;        case 2:movement(2); break;        case 3:movement(3); break;        case 4:movement(4); break;        }        if (life) pp();    }    dead();//shu chu si wang     return 0;}