Bitmap

package com.qdsoftware.snake.map;




import android.graphics.Bitmap;
import android.graphics.Canvas;
import android.graphics.Paint;
import android.graphics.Rect;


public class Map {
private int x;
private int y;

private int width;
private int height;

public int getX() {
return x;
}


public void setX(int x) {
this.x = x;
}


public int getY() {
return y;
}


public void setY(int y) {
this.y = y;
}
private Bitmap bg;
private Bitmap ww;
public Map(Bitmap bg,Bitmap ww,int width,int height){
this.bg = bg;
this.width = width;
this.height = height;
this.x = 0;
this.y = 0;
this.ww = ww;
}

private Paint p = new Paint();
public void onDraw(Canvas canvas){
for(int i = 0;i < 16;i++){
for(int j = 0;j < 16;j++){
int data = nMapData0[i][j];
int mapData = data&0xff;
if(mapData == 0){
continue;
}
//获取单元格的数组,并判断是那个图片
short[] bitmapData = nDrawPos[mapData - 1];
//新建Rect矩形,用来显示单元格
Rect src = new Rect();//src 这个是表示绘画图片在原图上的大小位置
Rect dst = new Rect();//dst 是表示 绘画这个图片在需要绘制的屏幕上的位置
if(bitmapData[0] == 0){
src.left = bitmapData[1];//bitmapData[] = {0,224,248,24,24};
src.top = bitmapData[2];
src.right = src.left + bitmapData[3];
src.bottom = src.top + bitmapData[4];
//确定绘制在屏幕上的位置
dst.left = bitmapData[3] * j;
dst.top = bitmapData[4] * i;
dst.right = dst.left + bitmapData[3];
dst.bottom = dst.top + bitmapData[4];
//绘制
canvas.drawBitmap(bg, src, dst, p);
}else if(bitmapData[0] == 1){
src.left = bitmapData[1];//bitmapData[] = {1,223,161,24,24};
src.top = bitmapData[2];
src.right = src.left + bitmapData[3];
src.bottom = src.top + bitmapData[4];
//确定绘制在屏幕上的位置
dst.left = 24 * j;
dst.top = 24 * i;
dst.right = dst.left + bitmapData[3];
dst.bottom = dst.top + bitmapData[4];
//绘制
canvas.drawBitmap(ww, src, dst, p);
}
}
}

}

public void onLogic(){
if(y==height){
y=0;
return;
}
y+=10;
}

//子图片索引表:{图片,左,上,宽,高,[旋转及镜象]}
short nDrawPos[][]={
{0,97,104,24,24},   //索引:[0]
{1,214,164,24,24},  //索引:[1]
{0,0,0,24,24},      //索引:[2]
{0,498,108,24,24}   //索引:[3]
};
//地图数据16x16
int nMapData0[][]={
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0101,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0101,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0101,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0101,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0002,0x0001,0x0001,0x0101,0x0001},
{0x0001,0x0001,0x0002,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001},
{0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001,0x0001}};







}
下载方式:https://pan.quark.cn/s/a4b39357ea24 布线问题(分支限界算法)是计算机科学和电子工程领域中一个广为人知的议题,它主要探讨如何在印刷电路板上定位两个节点间最短的连接路径。 在这一议题中,电路板被构建为一个包含 n×m 个方格的矩阵,每个方格能够被界定为可通行或不可通行,其核心任务是定位从初始点到最终点的最短路径。 分支限界算法是处理布线问题的一种常用策略。 该算法与回溯法有相似之处,但存在差异,分支限界法仅需获取满足约束条件的一个最优路径,并按照广度优先或最小成本优先的原则来探索解空间树。 树 T 被构建为子集树或排列树,在探索过程中,每个节点仅被赋予一次成为扩展节点的机会,且会一次性生成其全部子节点。 针对布线问题的解决,队列式分支限界法可以被采用。 从起始位置 a 出发,将其设定为首个扩展节点,并将与该扩展节点相邻且可通行的方格加入至活跃节点队列中,将这些方格标记为 1,即从起始方格 a 到这些方格的距离为 1。 随后,从活跃节点队列中提取队首节点作为下一个扩展节点,并将与当前扩展节点相邻且未标记的方格标记为 2,随后将这些方格存入活跃节点队列。 这一过程将持续进行,直至算法探测到目标方格 b 或活跃节点队列为空。 在实现上述算法时,必须定义一个类 Position 来表征电路板上方格的位置,其成员 row 和 col 分别指示方格所在的行和列。 在方格位置上,布线能够沿右、下、左、上四个方向展开。 这四个方向的移动分别被记为 0、1、2、3。 下述表格中,offset[i].row 和 offset[i].col(i=0,1,2,3)分别提供了沿这四个方向前进 1 步相对于当前方格的相对位移。 在 Java 编程语言中,可以使用二维数组...
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