C++基于easyx怎么实现迷宫游戏

本篇内容介绍了"C++基于easyx怎么实现迷宫游戏"的有关知识，在实际案例的操作过程中，不少人都会遇到这样的困境，接下来就让小编带领大家学习一下如何处理这些情况吧！希望大家仔细阅读，能够学有所成！

效果：

/*走迷宫*/#define _CRT_SECURE_NO_DEPRECATEd#define _CRT_SECURE_NO_WARNINGS#include#include#include#include#define LEFT            0//方向#define RIGHT            1#define UP            0//由于当前素材只有左右二个方向，所以上下共用了左右方向#define DOWN            1#define ROAD 0//地图元素类型#define WALL 1#define ENTERX 1//入口  x列，y行#define ENTERY 0#define OUTX 11 //出口 x列，y行#define OUTY 8#define HUMANWIDTH        75#define HUMANHEIGHT        130#define WIDTH            12//地图大小#define HEIGHT            10IMAGE img_human;IMAGE img_human_mask;IMAGE img_wall;IMAGE img_road;int moveNum[2] = { 0 };//当前动作序号int direction;//上下左右四个方向int human_witdh;int human_height;int x, y;//x列数，y行数int map[HEIGHT][WIDTH] = {//地图    { 1,1,1,1,1,1,1,1,1,1,1,1 },    { 0,0,0,1,1,1,1,1,1,1,1,1 },    { 1,1,0,1,1,1,1,0,1,1,1,1 },    { 1,1,0,0,1,1,1,0,1,1,1,1 },    { 1,1,1,0,1,1,1,0,1,1,1,1 },    { 1,1,1,0,1,1,1,0,1,1,1,1 },    { 1,1,1,0,1,1,1,0,1,1,1,1 },    { 1,1,1,0,0,0,0,0,0,0,1,1 },    { 1,1,1,1,1,1,1,1,1,0,0,0 },    { 1,1,1,1,1,1,1,1,1,1,1,1 },};void showbk() {//绘制背景    for (int j = 0; j < WIDTH; j++)        for (int i = 0; i < HEIGHT; i++)            if (map[i][j] == WALL)                putimage(j * img_wall.getwidth(), i * img_wall.getheight(), img_wall.getwidth(), img_wall.getheight(), &img_wall, 0, 0, SRCCOPY);            else putimage(j * img_wall.getwidth(), i * img_wall.getheight(), img_wall.getwidth(), img_wall.getheight(), &img_road, 0, 0, SRCCOPY);}void start()//初始化{    loadimage(&img_wall, _T(".\\walls.gif"));    initgraph(img_wall.getwidth() * WIDTH, img_wall.getheight() * HEIGHT);    loadimage(&img_human, _T(".\\行走素材图.jpg"));    loadimage(&img_human_mask,_T( ".\\行走素材图mask.jpg"));    human_witdh = 75;//img_human.getwidth()/4;    human_height = 130;//img_human.getheight()/2;                       //putimage(x,y,HUMANWIDTH,HUMANHEIGHT,&img_human,0,0);    loadimage(&img_road, _T(".\\road.gif"));    x = 0;    y = 1;}void updateWithoutInput(){}void drawRole(int x0, int y0)//绘制前景{    putimage((x - x0 / 4.0) * img_wall.getwidth() - 7,        (y - y0 / 4.0) * img_wall.getheight() - 70,        human_witdh, human_height, &img_human_mask, moveNum[direction] * human_witdh, direction * (human_height - 10), NOTSRCERASE);    putimage((x - x0 / 4.0) * img_wall.getwidth() - 7,        (y - y0 / 4.0) * img_wall.getheight() - 70,        human_witdh, human_height, &img_human, moveNum[direction] * human_witdh, direction * (human_height - 10), SRCINVERT);}void show(int x0, int y0){    showbk();    //clearrectangle(x,y,x+human_witdh,y+human_height);        //先显示背景    //准备好遮罩MASK图和源图，三元光栅操作    drawRole(x0, y0);    FlushBatchDraw();    Sleep(50);}void readRecordFile(){//读取存档    FILE* fp;    int temp;    fp = fopen(".\\record.dat", "r");    fscanf(fp, "%d %d", &x, &y);    fclose(fp);}void WriteRecordFile(){//保存存档    FILE* fp;    int temp;    fp = fopen(".\\record.dat", "w");    fprintf(fp, "%d %d ", x, y);    fclose(fp);}void updateWithInput(){//增加过度    char input;    int olddirection = direction;    int oldx = x;    int oldy = y;    /******异步输入检测方向键状态    if(GetAsyncKeyState(VK_LEFT)&0x8000)  向左    if(GetAsyncKeyState(VK_RIGHT)&0x8000)  向右    if(GetAsyncKeyState(VK_UP)&0x8000)  向上    if(GetAsyncKeyState(VK_DOWN)&0x8000)  向下    ********/    if (_kbhit())    {        input = _getch();        switch (input)        {        case 'a':direction = LEFT;        if (map[y][x - 1] == ROAD) x--; moveNum[direction] = 0; break;        case 'd':direction = RIGHT;        if (map[y][x + 1] == ROAD) x++; moveNum[direction] = 0; break;        case 'w':direction = UP;        if (map[y - 1][x] == ROAD) y--; moveNum[direction] = 0; break;        case 's':direction = DOWN;        if (map[y + 1][x] == ROAD) y++; moveNum[direction] = 0; break;        case 'W':WriteRecordFile(); break;        case 'R':readRecordFile(); break;        }        if (x != oldx || y != oldy)            for (int i = 4; i > 0; i--)            {//过渡动画                show((x - oldx) * i, (y - oldy) * i);                moveNum[direction]++;//动作序号，一个完整动作分解为四个姿势                moveNum[direction] %= 4;            }    }}int main(){    start();    BeginBatchDraw();    while (1) {        show(0, 0);        Sleep(50);        if (x == OUTX && y == OUTY)//到达了出口        {            outtextxy(0, 0, _T("reach target!"));            Sleep(50);            break;        }        updateWithoutInput();        updateWithInput();    }    EndBatchDraw();    _getch();    closegraph();    return 0;}

"C++基于easyx怎么实现迷宫游戏"的内容就介绍到这里了，感谢大家的阅读。如果想了解更多行业相关的知识可以关注网站，小编将为大家输出更多高质量的实用文章！