第一遍看作业要求没看懂。。不过看懂了后算法实现起来还是蛮简单的，主要在于遍历每个墙时，找出其分隔的两个元素，若两元素间无path，把墙拆了2333，有path的话墙不拆，注意那个它给定union(root1,root2)函数接受的两个参数必须是root，所以正确的写法应该是union(find(i),find(j))，若传入了非root的元素会产生死循环。
代码：
1 private class Walls{ 2 int h; 3 int v; 4 boolean isHoriz; 5 Walls(int a,int b,boolean c){ 6 h=a; 7 v=b; 8 isHoriz=c; 9 } 10 } 11 public Maze(int horizontalSize, int verticalSize) { 12 int i, j; 13 14 horiz = horizontalSize; 15 vert = verticalSize; 16 if ((horiz < 1) || (vert < 1) || ((horiz == 1) && (vert == 1))) { 17 return; // There are no interior walls 18 } 19 20 // Create all of the horizontal interior walls. Initially, every 21 // horizontal wall exists; they will be removed later by the maze 22 // generation algorithm. 23 if (vert > 1) { 24 hWalls = new boolean[horiz][vert - 1]; 25 for (j = 0; j < vert - 1; j++) { 26 for (i = 0; i < horiz; i++) { 27 hWalls[i][j] = true; 28 } 29 } 30 } 31 // Create all of the vertical interior walls. 32 if (horiz > 1) { 33 vWalls = new boolean[horiz - 1][vert]; 34 for (i = 0; i < horiz - 1; i++) { 35 for (j = 0; j < vert; j++) { 36 vWalls[i][j] = true; 37 } 38 } 39 } 40 int wallnumbers=horiz*(vert-1)+vert*(horiz-1); 41 Walls[]walls=new Walls[wallnumbers]; 42 int count=0; 43 for( i=0;i<horiz;i++){ 44 for( j=0;j<vert-1;j++){ 45 walls[count]=new Walls(i,j,true); 46 count++; 47 } 48 } 49 for(i=0;i<horiz-1;i++){ 50 for(j=0;j<vert;j++){ 51 walls[count]=new Walls(i,j,false); 52 count++; 53 } 54 } 55 for(i=wallnumbers-1;i>0;i--){ 56 int r=randInt(i); 57 Walls template=walls[i]; 58 walls[i]=walls[r]; 59 walls[r]=template; 60 } 61 DisjointSets set=new DisjointSets(this.horiz*this.vert); 62 try{ 63 for(i=0;i<wallnumbers;i++){ 64 if(walls[i].isHoriz){ 65 int number1=walls[i].h+walls[i].v*horiz; 66 int number2=walls[i].h+(walls[i].v+1)*horiz; 67 if(set.find(number1)==set.find(number2)){ 68 hWalls[walls[i].h][walls[i].v]=true; 69 } 70 else{ 71 hWalls[walls[i].h][walls[i].v]=false; 72 set.union(set.find(number1),set.find(number2)); 73 } 74 } 75 else if(!walls[i].isHoriz){ 76 int number1=walls[i].h+walls[i].v*horiz; 77 int number2=number1+1; 78 if(set.find(number1)==set.find(number2)) 79 vWalls[walls[i].h][walls[i].v]=true; 80 else{ 81 vWalls[walls[i].h][walls[i].v]=false; 82 set.union(set.find(number1), set.find(number2)); 83 } 84 } 85 } 86 }catch(Exception e){ 87 e.printStackTrace(); 88 }
运行结果：
一个10*10的maze:
---------------------
| | | | | | |
+ +-+ + + +-+ + +-+-+
| | | | | | |
+ +-+ +-+ +-+ + + +-+
| | | |
+-+ + +-+-+-+ +-+ +-+
| | | | |
+ +-+ + +-+ + + + +-+
| | | | | | | | |
+-+ +-+ +-+-+-+ + +-+
| | | | |
+-+ + +-+ + +-+-+ + +
| | | | | | | |
+ + + + + + +-+-+ +-+
| | | | | | | |
+ +-+ +-+ +-+ + + +-+
| | | | |
+-+ +-+ + +-+-+ + + +
| | | | |
+--------------------
What a fine maze you've created!
part2 那个没太搞懂啊，如果用DFS的意思是不是每次搜到一层的话，随机选一个没有visited的edge，然后其他的edge全无效，这样保障两个vertice之间只有一条path？