迷宫游戏C语言实现

//

//  main.c

//  444

//

//  Created by 安广沛 on 15/8/14.

//  Copyright (c) 2015年 agp. All rights reserved.

//

//##########

//#x########

//#     ####

//# ### ####

//# ###    #

//# ########

//# ####

//#      ###

//##########

//for (int i = 0; i< 9 ; i++) {

//    for (int j= 0; j<10; j++) {

//        printf("%c",*(*(p+i)+j));

//    }

//    printf("\n");

//}

#include <stdio.h>

void daYinDiTu(char (* p)[11]){

    

    

        for (int j= 0; j < 9; j++) {

            printf("%s",p[j]);

      

            printf("\n");}

   

    

    

}

//p=a=&a[0]

int main(int argc, const char * argv[]) {

    char diTu[9][11]=

    

    {

        "##########",

        "#x########",

        "#     ####",

        "# ### ####",

        "# ###    #",

        "# ########",

        "# ####    ",

        "#      ###",

        "##########"

    };

    //    char *p[]= {

    //        "##########",

    //        "#x########",

    //        "#     ####",

    //        "# ### ####",

    //        "# ###    #",

    //        "# ########",

    //        "# ####    ",

    //        "#      ###",

    //        "##########"

    //    };

    

    //打印地图

    daYinDiTu(diTu);

    //提示用户,W,S,A,D分别控制方向上下左右

    printf("W,S,A,D分别控制方向上下左右   \n");

    //定义变量接收用户所输入的方向

    char fangXiang;

    int x1=1,x2=1,y1=1,y2=1;//x(0,8),y(0,9)

    //用户输入方向

    while (1) {

        // 接受用户输入的方向

        scanf("%c",&fangXiang);

        getchar();// 接收回车

        switch (fangXiang) {

            case 'w':

            case 'W':

                x2 = x1-1;

                if (diTu[x1-1][y2]==' ') {

                    

                    int f;

                    f = diTu[x1][y1];

                    diTu[x1][y1] = diTu[x2][y2];

                    diTu[x2][y2] = f;

                    x1 = x2;

                }else

                {

                    x2 = x1;

                

                

                }

                

                daYinDiTu(diTu);

                

                

                break;

            case 's':

            case 'S':

                 x2 = x1+1;

                

                if (diTu[x1+1][y2] == ' ') {

                   

                    int f;

                    f = diTu[x1][y1];

                    diTu[x1][y1] = diTu[x2][y2];

                    diTu[x2][y2] = f;

                    x1 = x2;

                }else

                {

                    x2 = x1;

                    

                    

                }

                daYinDiTu(diTu);

                break;

            case 'A':

            case 'a':

                

                y2 = y1-1;

                if (diTu[x2][y1-1] == ' ') {

                    

                    int f;

                    f = diTu[x1][y1];

                    diTu[x1][y1] = diTu[x2][y2];

                    diTu[x2][y2] = f;

                    y1 = y2;

                }else{

                

                

                

                    y2 = y1;

                

                }

                daYinDiTu(diTu);

                break;

            case 'd':

            case 'D':

                y2 = y1 + 1;

                

                if (diTu[x2][y1+1] == ' ') {

                    

                    int f;

                    f = diTu[x1][y1];

                    diTu[x1][y1] = diTu[x2][y2];

                    diTu[x2][y2] = f;

                    y1 = y2;

                }else{

                    

                    

                    

                    y2 = y1;

                    

                }

                daYinDiTu(diTu);

                break;

                // Y = 9时

            default:

                printf("输入无效方向");

                break;

                

        }

        if (y1 == 9) {

            printf("恭喜你走出迷宫\n");

            return 0;

        }

    }

    

    //定义变量记录X当前的位置

    //当y2等于10时,提示用户走出迷宫

    //当遇到墙时,continue跳出当前循环

    

    

    

    

    

    

    return 0;

}