瓦片地图游戏开发的底层代码

原理：

二维数组存储每个瓦片序号

然后有一个缓冲区存储瓦片图片，

最后连续采样缓冲区，粘贴到屏幕上，

而缓冲区数据随着采样越界再重新更新

#include 
#include 

// 默认游戏地图 
int map[20][20]= {
	{0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,1,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,1,0,0, 0,0,2,0,2, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,0,0,2, 0,0,0,2,0, 0,0,0,0,0},
	{0,0,0,0,1, 0,0,0,0,0, 0,0,1,0,0, 1,0,0,0,0},
	
	{0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,1,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,1,0,0, 0,0,2,0,2, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,0,0,2, 0,0,0,2,0, 0,0,0,0,0},
	{0,0,0,0,1, 0,0,0,0,0, 0,0,1,0,0, 1,0,0,0,0},
	
	{0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,1,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,1,0,0, 0,0,2,0,2, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,0,0,2, 0,0,0,2,0, 0,0,0,0,0},
	{0,0,0,0,1, 0,0,0,0,0, 0,0,1,0,0, 1,0,0,0,0},
	
	{0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,1,0,0,0, 0,0,0,0,0, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,1,0,0, 0,0,2,0,2, 0,0,0,0,0},
	{0,0,0,0,0, 0,0,0,0,2, 0,0,0,2,0, 0,0,0,0,0},
	{0,0,0,0,1, 0,0,0,0,0, 0,0,1,0,0, 1,0,0,0,0},
};
int a=30;
int xc=3*a;
int yc=3*a;
IMAGE c(3*a,3*a);
int see = 9;
int seex =0;
int seey=0;
IMAGE n(see*a,see*a);
int oldnx=0;
int oldny=0;

void loadmap();

void change(ExMessage msg) {
	if(msg.message==WM_KEYDOWN) {
		switch(msg.vkcode) {
			case 0x41:
				xc--;
				break;
			case 0x44:
				xc++;
				break;
			case 0x57:
				yc--;
				break;
			case 0x53:
				yc++;
				break;
		}
	}
	int changeflag=0;
	if(xc>2*(3*a)) {
		xc=xc-(3*a);
		seex+=3;
		changeflag=1;
	}

	if(changeflag) {
		SetWorkingImage(&n);
		for(int i=0; i<9; i++) {
			for(int j=0; j<9; j++) {
				switch (map[i+seey][j+seex]) {				// 解决x坐标与数组的对应关系 第二个方括号【】是列数，是横坐标x,控制第几列
					case 0:
						setfillcolor(BLACK);
						break;
					case 1:
						setfillcolor(WHITE);
						break;
					case 2:
						setfillcolor(GREEN);
						break;
					case 3:
						setfillcolor(BLUE);
						break;
				}
				fillrectangle(j*a,i*a,(j+1)*a,(i+1)*a);
			}
		}
	}

	SetWorkingImage(&n);
	getimage(&c,xc,yc,3*a,3*a);
	SetWorkingImage();
	putimage(800,a,&n);
	putimage(1300,a,&c);
	printf("cx = %d, yc = %d\n",xc,yc);
	printf("seex = %d, seey = %d\n",seex,seey);
}


int main() {
	loadmap();
	getimage(&n,1*a,1*a,see*a,see*a);
	putimage(800,a,&n);
	ExMessage msg;
	while(1) {
		peekmessage(&msg,EX_KEY|EX_MOUSE,true);
		change(msg);
		Sleep(2);
		flushmessage(EX_MOUSE);
	}
	closegraph();
	return 0;
}



void loadmap() {
//	FILE* fp;
//	fp=fopen("map.txt","r");
//	for(int i=0; i<20; i++) {
//		for(int j=0; j<20; j++) {
//			fscanf(fp," %d",&map[i][j]);
//		}
//		fscanf(fp,"\n");
//	}
//
//	printf("gammap\n");
//	for(int i=0; i<20; i++) {
//		for(int j=0; j<20; j++) {
//			printf(" %d",map[i][j]);
//		}
//		printf("\n");
//	}
//	fclose(fp);

	initgraph(1800,800,1);
	setbkcolor(GREEN);
	cleardevice();

	for(int i=0; i<20; i++) {
		for(int j=0; j<20; j++) {
			rectangle(i*a,j*a,(i+1)*a,(j+1)*a);
		}
	}

	setfillcolor(WHITE);
	for(int i=0; i<20; i++) {
		for(int j=0; j<20; j++) {
			switch (map[i][j]) {
				case 0:
					setfillcolor(BLACK);
					break;
				case 1:
					setfillcolor(WHITE);
					break;
				case 2:
					setfillcolor(GREEN);
					break;
				case 3:
					setfillcolor(BLUE);
					break;
			}
			fillrectangle(j*a,i*a,(j+1)*a,(i+1)*a);
		}
	}
}