不相交類集演算法生成迷宮並求解路徑
阿新 • • 發佈:2018-11-10
這兩天看了不相交類集演算法,十分高效於是寫了一個迷宮生成的演算法,事實證明該演算法效率極高;但求解路徑遇到了瓶頸,樓主這裡用鄰接表的方式尋找路徑,用棧儲存路徑資訊。該程式存在一個小問題:pain()函式閃爍的導致多次計算;而且路徑求解的演算法效率不高,本人菜鳥一枚,希望路過的大神予以指點。
public final class DisJoinSet {
private int[] eleRoots;
public DisJoinSet(int num){
this.eleRoots = new int[num];
for(int i=0;i<num;i++){
getEleRoots()[i] = -1;
}
}
public int find(int ele){
if(getEleRoots()[ele] < 0){
return ele;
}
return find(getEleRoots()[ele]);
}
public void union(int root1,int root2){
//讓深度較小的樹成為深度較大的樹的子樹
if (getEleRoots()[root1] > getEleRoots()[root2]){
getEleRoots()[root1] = root2;
}else{
if(getEleRoots()[root1] == getEleRoots()[root2]){//深度一樣,則更新深度
getEleRoots()[root1]--;
}
getEleRoots()[root2] = root1;
}
}
public int[] getEleRoots() {
return eleRoots;
}
}
import javax.swing.*;
import java.awt.*;
import java.util.*;
import java.util.List;
/**
* Created by 偉大的華仔 on 2017-11-30.
*/
public class Maze extends JFrame {
private int row;//行數
private int col; //列數
private DisJoinSet disjSet;
private int winHeight=1000;// 行高
private int winWidth=1000;//行寬
public Maze(int row,int col){
this.row = row;
this.col = col;
this.setTitle("迷宮");//設定標題
this.setSize(winWidth,winHeight);//設定元件大小
this.setVisible(true);//顯示或隱藏此 Window
this.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);//設定使用者在此窗體上發起 "close" 時預設執行的操作 EXIT_ON_CLOSE(在 JFrame 中定義):使用 System exit 方法退出應用程式
}
public static void main(String[] args) {
/**
* 定義迷宮的大小,必須是正方形的
* */
int rowCount = 100;
int colCount = 100;
Maze maze = new Maze(rowCount,colCount);
}
@Override
public void paint(Graphics g){
super.paint(g);
/**
* 申請一個String集合,放置合併的節點
* */
List<Integer>DisList = new ArrayList<>();
//背景為白色
g.setColor(Color.white);
g.fillRect(0, 0, winWidth, winHeight);//視窗填充矩形
g.setColor(Color.black);
final int extraWidth = 20;
final int cellWidth = (winWidth-2*extraWidth)/row;//定義每個格子的寬度
final int cellHeight = (winHeight-4*extraWidth)/col;//定義每個格子的高度
for(int i=0;i<row;i++) {
for(int j=0;j<col;j++)
{
//初始化m*n矩陣格子
g.drawRect(i*cellWidth+extraWidth,j*cellHeight+2*extraWidth, cellWidth, cellHeight);
}
}
int lastPos = getLastElePos();//迷宮最後一個格式的代表數字
//起點,終點特殊處理
g.setColor(Color.red);
g.fillRect(extraWidth, 2*extraWidth, cellWidth, cellHeight);
g.fillRect((lastPos% row)*cellWidth + extraWidth,(lastPos/ row)*cellHeight + 2*extraWidth, cellWidth, cellHeight);
this.setDisjSet(new DisJoinSet(row*col));
// 路徑記錄操作
long start = System.currentTimeMillis();
ArrayList<LinkedList<Integer>> adjoinTable = new ArrayList();
LinkedList[] arrayFuck = new LinkedList[row*col];
for (int i = 0;i<row*col;i++){
LinkedList<Integer> temp = new LinkedList<Integer>();
temp.add(i);
arrayFuck[i] = temp;
temp.clear();
}
Stack<Integer> pathSearch = new Stack<Integer>();
pathSearch.push(0);
// for (int i = 0;i<row*col;i++){
// LinkedList<Integer> temp = new LinkedList<Integer>();
// temp.add(i);
// adjoinTable.add(temp);
// temp.clear();
// }
long end1 = System.currentTimeMillis();
System.out.println("構建資料消耗:"+(end1-start)+"ms");
g.setColor(Color.white); //用後景色擦色
while(disjSet.find(0) != disjSet.find(lastPos)){//如果起點和終點還沒在同一個等價類
/*
* 在迷宮內隨機挖一個點,再找到該點周圍一點,使這兩個點落在同一個等價類
*/
Random random = new Random();
int randPos = random.nextInt(lastPos+1);//+1是為了能隨機到最後一位
int rowIndex = randPos % row;
int colIndex = randPos / col;
List<Integer> neighborPos = getNeighborNums(rowIndex, colIndex) ;
int randNeighbor = neighborPos.get(random.nextInt(neighborPos.size()));
if(disjSet.find(randPos) == disjSet.find(randNeighbor)){//兩點在同一個等價類
continue;
}else{
int aRoot = disjSet.find(randPos);
int bRoot = disjSet.find(randNeighbor);
disjSet.union(aRoot, bRoot);
DisList.add(randPos);
DisList.add(randNeighbor);
// /**
// * 列印搜尋過程
// * */
// System.out.println("***過程分析*****************"+Arrays.toString(this.disjSet.getEleRoots()));
int maxNum = Math.max(randPos, randNeighbor);//取得較大點
int x1=0,y1=0,x2=0,y2=0;
if(Math.abs(randPos-randNeighbor) == 1){//說明在同一行,用豎線隔開
x1= x2=(maxNum% row)*cellWidth + extraWidth;
y1=(maxNum/ row)*cellHeight + 2*extraWidth;
y2=y1+cellHeight;
}else{//說明在同一列,用橫線隔開
y1=y2=(maxNum/ row)*cellHeight + 2*extraWidth;
x1=(maxNum% row)*cellWidth + extraWidth;
x2=x1+cellWidth;
}
g.drawLine(x1, y1, x2, y2);
}
}
long end2 = System.currentTimeMillis();
System.out.println("畫線和記錄資料資料消耗:"+(end2-end1)+"ms");
/**
* 列印陣列
* */
// System.out.println("***最終結構********************"+Arrays.toString(this.disjSet.getEleRoots()));
// for (Iterator it = DisList.iterator();it.hasNext();){
// System.out.println("***節點***"+it.next());
// }
for (int i = 0;i<DisList.size();i=i+2){
Integer j = DisList.get(i);
Integer k = DisList.get(i+1);
arrayFuck[j].addLast(k);
arrayFuck[k].addLast(j);
// adjoinTable.get(j).addLast(k);
// adjoinTable.get(k).addLast(j);
}
// for (int i = 0;i<adjoinTable.size();i++){
// System.out.println("連結串列內容"+Arrays.toString(adjoinTable.get(i).toArray()));
// }
long end3 = System.currentTimeMillis();
System.out.println("構建鄰接表消耗:"+(end3-end2)+"ms");
/**
* 演算法步驟
* 1、檢索i節點的聯通節點
* 2、存在就放入棧中(前提為棧中沒有該元素),不存在彈出棧頂元素(刪除鄰接表中的資料),或者為終點元素結束
* 3、重複步驟1
* */
while (true){//while開始
// System.out.println("******棧中內容:"+Arrays.toString(pathSearch.toArray()));
if (pathSearch.isEmpty()){
System.out.println("棧為空,迷宮沒有路徑");
break;
}
if (pathSearch.peek()==(row*col-1)){
System.out.println("迷宮已破解");
break;
}
//adjoinTable.get(pathSearch.peek()).size()==1 && pathSearch.peek() != 0
if (arrayFuck[pathSearch.peek()].size() == 1 && pathSearch.peek() != 0){
Integer temp = pathSearch.pop();
if(pathSearch.isEmpty()){
System.out.println("棧為空,迷宮沒有路徑");
break;
}else {
// adjoinTable.get(pathSearch.peek()).remove(temp);
// adjoinTable.get(temp).remove(pathSearch.peek());
arrayFuck[pathSearch.peek()].remove(temp);
arrayFuck[temp].remove(pathSearch.peek());
}
}else {
// pathSearch.search(adjoinTable.get(pathSearch.peek()).get(0))!= -1
if (pathSearch.search(arrayFuck[pathSearch.peek()].get(0))!= -1 ){
if (arrayFuck[pathSearch.peek()].size()>1){
// pathSearch.push(adjoinTable.get(pathSearch.peek()).get(1));
Iterator<Integer> boy = arrayFuck[pathSearch.peek()].listIterator();
while (boy.hasNext()){
int girl = boy.next();
if (!pathSearch.contains(girl)){
pathSearch.push(girl);
}
}
}
}else {
Integer arrayMenber = Integer.parseInt(arrayFuck[pathSearch.peek()].get(0).toString());
pathSearch.push(arrayMenber);
}
}
}//while結束
long end4 = System.currentTimeMillis();
System.out.println("遍歷路徑表消耗:"+(end4-end3)+"ms");
/**
* 輸出路徑
* */
// System.out.println("迷宮路徑"+Arrays.toString(pathSearch.toArray()));
g.setColor(Color.BLUE);
while (!pathSearch.isEmpty()){
int boy = pathSearch.pop();
//左右
g.fillOval((boy% row)*cellWidth + cellWidth/4+extraWidth,(boy/ row)*cellHeight + cellHeight/4+2*extraWidth, cellWidth/2, cellHeight/2);
}
long end5 = System.currentTimeMillis();
System.out.println("畫路徑消耗:"+(end5-end4)+"ms");
long end = System.currentTimeMillis();
System.out.println("時間流程:"+Long.toString(end-start)+"ms");
adjoinTable.clear();
pathSearch.clear();
}
/**
* 取得目標座標點周圍四個有效點
*/
public List<Integer> getNeighborNums(int rowIndex,int colIndex){
List<Integer> neighborPos = new ArrayList<Integer>(4);
//右元素
if(isPointInMaze(rowIndex+1,colIndex)){
neighborPos.add(getCoordinateNum(rowIndex+1,colIndex));
}
//下元素
if(isPointInMaze(rowIndex,colIndex+1)){
neighborPos.add(getCoordinateNum(rowIndex,colIndex+1));
}
//左元素
if(isPointInMaze(rowIndex-1,colIndex)){
neighborPos.add(getCoordinateNum(rowIndex-1,colIndex));
}
//上元素
if(isPointInMaze(rowIndex,colIndex-1)){
neighborPos.add(getCoordinateNum(rowIndex,colIndex-1));
}
return neighborPos;
}
public int getLastElePos(){
return row*col-1;
}
public DisJoinSet getDisjSet() {
return disjSet;
}
public void setDisjSet(DisJoinSet disjSet) {
this.disjSet = disjSet;
}
/**
* 根據座標返回對應的值
* 例如在4*3矩陣,(0,0)返回0;(3,2)返回10
*/
public int getCoordinateNum(int x,int y){
return y*col + x;
}
/**
* 判斷給定座標是否在迷宮矩陣內
*/
public boolean isPointInMaze(int x,int y){
if(x < 0 || y < 0) return false;
return x < row && y <col;
}
}
這裡為100x100的結果展示,耗時最多5s。200x200的迷宮大致需要20s左右。