摘要：本人郵箱歡迎轉(zhuǎn)載轉(zhuǎn)載請(qǐng)注明網(wǎng)址代碼已經(jīng)全部托管有需要的同學(xué)自行下載引言迷宮對(duì)于大家都不會(huì)陌生那么迷宮是怎么生成已經(jīng)迷宮要如何找到正確的路徑呢用代碼又怎么實(shí)現(xiàn)帶著這些問(wèn)題我們繼續(xù)往下看并查集朋友圈有一種算法就做并查集什么意思呢比如現(xiàn)在有零
本人郵箱:
歡迎轉(zhuǎn)載,轉(zhuǎn)載請(qǐng)注明網(wǎng)址 http://blog.csdn.net/tianshi_kco
github: https://github.com/kco1989/kco
代碼已經(jīng)全部托管github有需要的同學(xué)自行下載
引言
迷宮對(duì)于大家都不會(huì)陌生.那么迷宮是怎么生成,已經(jīng)迷宮要如何找到正確的路徑呢?用java代碼又怎么實(shí)現(xiàn)?帶著這些問(wèn)題.我們繼續(xù)往下看.
并查集(find-union)
朋友圈
有一種算法就做并查集(find-union).什么意思呢?比如現(xiàn)在有零散的甲乙丙丁戊五個(gè)人.他們之間剛開(kāi)始互相不認(rèn)識(shí).用代碼解釋就是find(person1, person2) == false,之后在某一次聚合中,甲認(rèn)識(shí)了乙,乙認(rèn)識(shí)了丙,丙認(rèn)識(shí)了戊和丁等等,那么就可以用代碼解釋如下.
union("甲", "乙");
union("乙", "丙");
union("丙", "戊");
union("丙", "丁");
那么這個(gè)時(shí)候甲就可以通過(guò)乙認(rèn)識(shí)到了丙在通過(guò)丙認(rèn)識(shí)到庚和丁.
這是甲乙丙丁戊通過(guò)朋友或者朋友的朋友最終都互相認(rèn)識(shí).換另一種說(shuō)法就是如果甲要認(rèn)識(shí)丁,那么甲必須先通過(guò)乙認(rèn)識(shí)丙,再通過(guò)丙去認(rèn)識(shí)丁就行了.
迷宮
對(duì)于迷宮生成,其實(shí)更上面朋友圈有點(diǎn)類似.生成步驟如下
首先,先創(chuàng)建一個(gè)n*m的二維密室.每個(gè)單元格四方都是墻.
隨機(jī)選擇密室中的一個(gè)單元格.之后在隨機(jī)選擇一面要打通的墻壁.
判斷要打通的墻壁是否為邊界.是則返回步驟3,不是則繼續(xù)
判斷步驟的單元個(gè)和要打通的墻壁的對(duì)面是否聯(lián)通(用find算法)
如果兩個(gè)單元格不聯(lián)通,則把步驟2選中的墻壁打通(用union算法).否則返回步驟2.
判斷迷宮起點(diǎn)和終點(diǎn)是否已經(jīng)聯(lián)通,是則迷宮生成結(jié)束,否則返回步驟2.
對(duì)于并查集(find-union)的實(shí)現(xiàn),網(wǎng)絡(luò)上有不少實(shí)現(xiàn),這里不展開(kāi)說(shuō)明了.
實(shí)現(xiàn)代碼
墻壁
public enum Wall {
/**
* 墻壁
*/
BLOCK,
/**
* 通道
*/
ACCESS
}
墻壁,是一個(gè)枚舉變量,用于判斷當(dāng)前的墻壁是否可以通行.
單元格
public class Box {
private Wall[] walls;
public Box(){
walls = new Wall[4];
for (int i = 0; i < walls.length; i ++){
walls[i] = Wall.BLOCK;
}
}
public void set(Position position, Wall wall){
walls[position.getIndex()] = wall;
}
public Wall get(Position position){
return walls[position.getIndex()];
}
}
一個(gè)單元格(小房間)是有四面墻組成的,剛開(kāi)始四面墻都是墻壁.
方向
public enum Position {
TOP(0), RIGHT(1), DOWN(2), LEFT(3);
public int index;
Position(int index) {
this.index = index;
}
public static Position indexOf(int index){
int pos = index % Position.values().length;
switch (pos){
case 0:
return TOP;
case 1:
return RIGHT;
case 2:
return DOWN;
case 3:
default:
return LEFT;
}
}
public Position anotherSide(){
switch (this){
case TOP:
return DOWN;
case RIGHT:
return LEFT;
case DOWN:
return TOP;
case LEFT:
default:
return RIGHT;
}
}
public int getIndex() {
return index;
}
}
方向,枚舉類,用于判斷單元格的那一面墻壁需要打通.
迷宮生成類
這里我使用find-union的兩種實(shí)現(xiàn)方式實(shí)現(xiàn),
一種是用數(shù)組的方式MazeArrayBuilder
一種使用map的方式實(shí)現(xiàn)MazeMapBuilder
所以我把迷宮生成的一些共同方法和屬性抽取出現(xiàn),編寫了一個(gè)抽象類AbstractMazeBuilder.然后再在MazeArrayBuilder或MazeMapBuilder做具體的實(shí)現(xiàn).
現(xiàn)在我們來(lái)看看AbstractMazeBuilder這個(gè)類
public abstract class AbstractMazeBuilder {
/**
* 迷宮行列最大值
*/
public static final int MAX_ROW_LINE = 200;
/**
* 行
*/
protected int row;
/**
* 列
*/
protected int line;
/**
* 迷宮格子集合,每個(gè)格子有四面墻
*/
protected Box[][] boxes;
/**
* 求解迷宮中間變量
*/
protected int[][] solverPath;
/**
* 迷宮時(shí)候已經(jīng)算出最佳路徑
*/
protected boolean isSolver;
/**
* 使用貪婪算法,算出最佳路徑集合
*/
protected List bestPath;
protected Random random;
public AbstractMazeBuilder(int row, int line){
if (row < 3 || row > MAX_ROW_LINE || line < 3 || line > MAX_ROW_LINE){
throw new IllegalArgumentException("row/line 必須大于3,小于" + MAX_ROW_LINE);
}
this.row = row;
this.line = line;
isSolver = false;
boxes = new Box[row][line];
solverPath = new int[row][line];
bestPath = new ArrayList();
random = new Random();
for (int i = 0; i < row; i ++){
for (int j = 0; j < line; j ++){
boxes[i][j] = new Box();
solverPath[i][j] = Integer.MAX_VALUE;
}
}
}
/**
* 查詢與point點(diǎn)聯(lián)通的最大格子的值
* @param point point
* @return 查詢與point點(diǎn)聯(lián)通的最大格子的值
*/
protected abstract int find(MazePoint point);
/**
* 聯(lián)通point1和point2點(diǎn)
* @param point1 point1
* @param point2 point2
*/
protected abstract void union(MazePoint point1, MazePoint point2);
/**
* 判斷時(shí)候已經(jīng)生成迷宮路徑
* @return 判斷時(shí)候已經(jīng)生成迷宮路徑
*/
protected abstract boolean hasPath();
/**
* 生成迷宮
*/
public void makeMaze(){
while (hasPath()){
// 生成 當(dāng)前點(diǎn), 當(dāng)前點(diǎn)聯(lián)通的方向, 當(dāng)前點(diǎn)聯(lián)通的方向?qū)?yīng)的點(diǎn)
ThreeTuple tuple = findNextPoint();
if (tuple == null){
continue;
}
union(tuple.one, tuple.three);
breakWall(tuple.one, tuple.two);
breakWall(tuple.three, tuple.two.anotherSide());
}
breakWall(new MazePoint(0,0), Position.LEFT);
breakWall(new MazePoint(row - 1, line - 1), Position.RIGHT);
}
/**
* 生成 當(dāng)前點(diǎn), 當(dāng)前點(diǎn)聯(lián)通的方向, 當(dāng)前點(diǎn)聯(lián)通的方向?qū)?yīng)的點(diǎn)
* @return
* ThreeTuple.one 當(dāng)前點(diǎn)
* ThreeTuple.two 當(dāng)前點(diǎn)聯(lián)通的方向
* ThreeTuple.three 當(dāng)前點(diǎn)聯(lián)通的方向?qū)?yīng)的點(diǎn)
*/
private ThreeTuple findNextPoint() {
MazePoint currentPoint = new MazePoint(random.nextInt(row), random.nextInt(line));
Position position = Position.indexOf(random.nextInt(Position.values().length));
MazePoint nextPoint = findNext(currentPoint, position);
if (nextPoint == null || find(currentPoint) == find(nextPoint)){
return null;
}
return new ThreeTuple(currentPoint, position, nextPoint);
}
/**
* 打通墻
* @param point 當(dāng)前點(diǎn)
* @param position 當(dāng)前點(diǎn)的方向
*/
private void breakWall(MazePoint point, Position position) {
boxes[point.x][point.y].set(position, Wall.ACCESS);
}
/**
* 通過(guò)當(dāng)前點(diǎn)以及對(duì)應(yīng)當(dāng)前點(diǎn)的方向找到下一個(gè)點(diǎn)
* @param currentPoint 當(dāng)前點(diǎn)
* @param position 方向
* @return 下個(gè)點(diǎn),若該點(diǎn)在迷宮內(nèi),這返回,否則返回null
*/
private MazePoint findNext(MazePoint currentPoint, Position position) {
MazePoint nextPoint;
switch (position){
case TOP:
nextPoint = new MazePoint(currentPoint.x - 1, currentPoint.y);
break;
case RIGHT:
nextPoint = new MazePoint(currentPoint.x, currentPoint.y + 1);
break;
case DOWN:
nextPoint = new MazePoint(currentPoint.x + 1, currentPoint.y);
break;
case LEFT:
default:
nextPoint = new MazePoint(currentPoint.x, currentPoint.y - 1);
break;
}
if (nextPoint.x < 0 || nextPoint.x >= row || nextPoint.y < 0 || nextPoint.y >= line){
return null;
}
return nextPoint;
}
public Box getBoxes(int x, int y) {
return boxes[x][y];
}
public int getRow() {
return row;
}
public int getLine() {
return line;
}
/**
* 求解迷宮路徑
* @return 迷宮路徑
*/
public List solvePath(){
// 1 迷宮時(shí)候已經(jīng)求解完成,是的話,則直接返回,不必再次計(jì)算
if (isSolver){
return bestPath;
}
// 2 否則計(jì)算迷宮最佳路徑
bestPath = new ArrayList();
solverPath(new MazePoint(0, 0), 0);
addPath(new MazePoint(row - 1, line - 1));
Collections.reverse(bestPath);
isSolver = true;
return bestPath;
}
/**
* 從終點(diǎn)逆推,添加最佳路徑
* @param point 當(dāng)前點(diǎn)
*/
private void addPath(MazePoint point) {
bestPath.add(point);
// 遍歷當(dāng)前點(diǎn)的每個(gè)方向,如果該方向能聯(lián)通,這步數(shù)跟當(dāng)前點(diǎn)的步數(shù)相差1步,這添加改點(diǎn),遞歸計(jì)算下去
for (Position position :Position.values()){
MazePoint next = findNext(point, position);
if (next == null || getBoxes(point.x, point.y).get(position) == Wall.BLOCK){
continue;
}
if (solverPath[point.x][point.y] - 1 == solverPath[next.x][next.y]){
addPath(next);
return;
}
}
}
/**
* 遞歸求解迷宮最佳路徑
* @param point 當(dāng)前點(diǎn)
* @param count 從開(kāi)始走到當(dāng)前點(diǎn)所需要的步數(shù)
*/
private void solverPath(MazePoint point, int count) {
// 判斷當(dāng)前點(diǎn)的步數(shù)時(shí)候小于現(xiàn)在走到這個(gè)點(diǎn)的步數(shù),
// 如果當(dāng)前點(diǎn)的步數(shù)比較小,則直接返回
if (solverPath[point.x][point.y] <= count){
return;
}
// 否則表示當(dāng)前點(diǎn),有更短的路徑
solverPath[point.x][point.y] = count;
// 再遍歷當(dāng)前點(diǎn)的每個(gè)方向
for (Position position : Position.values()){
MazePoint next = findNext(point, position);
// 如果下一個(gè)點(diǎn)不在迷宮內(nèi),或當(dāng)前點(diǎn)對(duì)應(yīng)的方向是一面墻壁,則跳過(guò)繼續(xù)編寫下一個(gè)方向
if (next == null || getBoxes(point.x, point.y).get(position) == Wall.BLOCK){
continue;
}
// 否則,步數(shù)加1, 遞歸計(jì)算
solverPath(next, count + 1);
}
}
public static class MazePoint{
public final int x;
public final int y;
public MazePoint(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public String toString() {
return "MazePoint{" +
"x=" + x +
", y=" + y +
"}";
}
}
}
代碼上有注釋,理解起來(lái)還是比較容易.MazeArrayBuilder和MazeMapBuilder的實(shí)現(xiàn)就參考github了.
AbstractMazeBuilder 中還包括了迷宮的求解.
迷宮的求解
迷宮的求解,一般我會(huì)使用以下兩種方法
右手規(guī)則,從起點(diǎn)出發(fā),遇到墻壁,則向右手邊轉(zhuǎn),按照這個(gè)規(guī)則.一般是可以找到出口的.不過(guò)如果迷宮有閉環(huán),則無(wú)法求解,而且解出來(lái)的路徑也不是最短路徑.
迷宮最短路徑算法.
從起點(diǎn)出發(fā),計(jì)數(shù)count=0
遍歷該點(diǎn)的任意方向,如果是墻壁,則忽略,不然count++,進(jìn)入下一個(gè)聯(lián)通的格子
判斷當(dāng)前格子的的count(默認(rèn)值一般是比較大的數(shù))是否比傳入的參數(shù)大,是說(shuō)明該格子是一條捷徑,將當(dāng)前各自的count=入?yún)?繼續(xù)第2步;否則,說(shuō)明該點(diǎn)已經(jīng)被探索過(guò)且不是一條捷徑,忽略
如果反復(fù),暴力遍歷所有單元格,即可以求出最短路徑
遍歷完之后,從出口開(kāi)始找,此時(shí)出口的數(shù)字,表示從入口走到出口需要的最小步數(shù).依次減1,找到下一個(gè)格子,直到找打入口.則最短路徑就生成了.
附加運(yùn)行結(jié)果
打賞
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