初学者的C语言智能蛇

单纯的书本知识的记忆不论对于哪门学科来说都是枯燥乏味的，而若能运用自己已学的知识去解决或创造一个问题，学习将会被赋予更为明显的意义。于是，在我们整天盯着书本而不知所以时，不如来段游戏设计吧。
若你不知从哪里开始着手，相信在屏幕上打一条可以自动寻路的贪吃蛇（我更愿称它为智能蛇）将会是一个不错的开始。而且倘若你有兴趣，这条蛇甚至可以陪伴你的整个学习历程，在你的手中越来越“强大”。

//这是源代码
#include<stdio.h>
#include<stdlib.h>
#include<windows.h>
#include"kky.h"
int main(void) {
    int io, oi;
    int m, n;
    int m0, n0;
    int a, b, c;
    int t = 0;
    int y = 1;
    int k[2] = { 4, 4 };
    for (io = 0; io < COUNT; io++) {
        for (oi = 0; oi < COUNT; oi++) {
            q[io][oi].w = 0;
            q[io][oi].e = ' ';
        }
    }
    for (io = 0; io < COUNT; io++) {
        for (oi = 0; oi < COUNT; oi++) {
            if (io == COUNT - 1 || io == 0 || oi == COUNT - 1 || oi == 0)
                q[io][oi].e = '*';
        }
    }
    m0 = area(COUNT);
    n0 = area(COUNT);
    q[m0][n0].e = 48;
    q[m0][n0].w = 1;
    a = m0;
    b = n0;
    while (q[a][b].e != ' ') {
        a = area(COUNT);
        b = area(COUNT);
    }
    m = m0;
    n = n0;
    q[a][b].e = 'x';
    t++;
    k[1] = news(&k[0], 4);
    while (y) {
        system("cls");
        while ((c = next(k[1], &m0, &n0)) != 4 || q[m0][n0].e == 'o') {
            m0 = m;
            n0 = n;
            k[1] = news(&k[0], c);
        }
        m = m0;
        n = n0;
        if (q[m][n].e == 'x') {
            t++;
            q[m][n].w = 1;
            for (io = 0; io < COUNT; io++) {
                for (oi = 0; oi < COUNT; oi++) {
                    if (q[io][oi].w != 0) {
                        q[io][oi].w++;
                        q[io][oi].e = 'o';
                    }
                }
            }
            q[m][n].w--;
            q[m][n].e = '0';
            while (q[a][b].e != ' ') {
                a = area(COUNT);
                b = area(COUNT);
            }
            q[a][b].e = 'x';
        }
        else {
            q[m][n].w = 1;
            for (io = 0; io < COUNT; io++) {
                for (oi = 0; oi < COUNT; oi++) {
                    if (q[io][oi].w != 0) {
                        q[io][oi].w++;
                        q[io][oi].e = 'o';
                    }
                    if (q[io][oi].w == t + 1) {
                        q[io][oi].w = 0;
                        q[io][oi].e = ' ';
                    }
                }
            }
            q[m][n].w = 1;
            q[m][n].e = '0';
        }
        for (io = 0; io < COUNT; io++) {
            for (oi = 0; oi < COUNT; oi++) {
                putchar(q[io][oi].e);
            }
            putchar('\n');
        }
        k[1] = news(&k[0], 4);
        if (q[m + 1][n].e != ' ' && q[m - 1][n].e != ' ' && q[m][n + 1].e != ' ' && q[m][n - 1].e != ' ')
            y = 0;
        Sleep(100);
    }
    puts("\n\n     game over\nyour score is %d \n", t - 1);
    return 0;
}

//这是头文件
#define COUNT 9
int news(int *, int);
int area(int);
int next(int, int *, int *);
void print(void);
struct Area {
    int w;
    char e;
} q[COUNT][COUNT];

//这是函数实现源文件
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
#include"kky.h"
int news(int *q, int t) {
    int p;
    int k = *q;
    if (k == 3)
        k = 1;
    else if (k == 0)
        k = 2;
    else if (k == 1)
        k = 3;
    else if (k == 2)
        k = 0;
    srand((unsigned)time(NULL));
    while ((p = rand() % 4) == k || p == t);
    *q = p;
    return p;
}

int area(int q) {
    srand((unsigned)time(NULL));
    return (rand() % (q - 2) + 1);
}

int next(int k, int *m, int *n) {
    switch (k) {
    case 0:
        if (*n + 1 >= COUNT - 1)
            return 0;
        (*n)++;
        break;
    case 1:
        if (*m + 1 >= COUNT - 1)
            return 1;
        (*m)++;
        break;
    case 2:
        if (*n - 1 <= 0)
            return 2;
        (*n)--;
        break;
    case 3:
        if (*m - 1 <= 0)
            return 3;
        (*m)--;
        break;
    }
    return 4;
}

void print(void) {
    int io, oi;
    for (io = 0; io < COUNT; io++) {
        for (oi = 0; oi < COUNT; oi++) {
            putchar(q[io][oi].e);
        }
        putchar('\n');
    }
}

（其中对贪吃蛇下一步方向的判断的实现上写得是很糟糕的，读者可以自己重新设计一下这方面的算法，因为人较懒就没打算优化了，不过应该很简单的。运行应该没什么问题。有兴趣可以玩一下。至于详细解析就不在这写了，想要的话可以翻一翻我的博客。我会在下一篇中介绍的。）