JAVA程序设计：推箱子（LeetCode：5266）

「推箱子」是一款风靡全球的益智小游戏，玩家需要将箱子推到仓库中的目标位置。

游戏地图用大小为 n * m 的网格 grid 表示，其中每个元素可以是墙、地板或者是箱子。

现在你将作为玩家参与游戏，按规则将箱子 'B' 移动到目标位置 'T' ：

玩家用字符 'S' 表示，只要他在地板上，就可以在网格中向上、下、左、右四个方向移动。
地板用字符 '.' 表示，意味着可以自由行走。
墙用字符 '#' 表示，意味着障碍物，不能通行。 
箱子仅有一个，用字符 'B' 表示。相应地，网格上有一个目标位置 'T'。
玩家需要站在箱子旁边，然后沿着箱子的方向进行移动，此时箱子会被移动到相邻的地板单元格。记作一次「推动」。
玩家无法越过箱子。
返回将箱子推到目标位置的最小 推动 次数，如果无法做到，请返回 -1。

示例 1：

输入：grid = [["#","#","#","#","#","#"],
             ["#","T","#","#","#","#"],
             ["#",".",".","B",".","#"],
             ["#",".","#","#",".","#"],
             ["#",".",".",".","S","#"],
             ["#","#","#","#","#","#"]]
输出：3
解释：我们只需要返回推箱子的次数。
示例 2：

输入：grid = [["#","#","#","#","#","#"],
             ["#","T","#","#","#","#"],
             ["#",".",".","B",".","#"],
             ["#","#","#","#",".","#"],
             ["#",".",".",".","S","#"],
             ["#","#","#","#","#","#"]]
输出：-1
示例 3：

输入：grid = [["#","#","#","#","#","#"],
             ["#","T",".",".","#","#"],
             ["#",".","#","B",".","#"],
             ["#",".",".",".",".","#"],
             ["#",".",".",".","S","#"],
             ["#","#","#","#","#","#"]]
输出：5
解释：向下、向左、向左、向上再向上。
示例 4：

输入：grid = [["#","#","#","#","#","#","#"],
             ["#","S","#",".","B","T","#"],
             ["#","#","#","#","#","#","#"]]
输出：-1
 

提示：

1 <= grid.length <= 20
1 <= grid[i].length <= 20
grid 仅包含字符 '.', '#',  'S' , 'T', 以及 'B'。
grid 中 'S', 'B' 和 'T' 各只能出现一个。

思路：对于箱子的移动我们采用深搜寻求最短路，每次箱子能移动的前提是人处在合适的位置上，因此我们对人的移动采用深搜进行模拟箱子移动。

class Solution {private boolean[][] flag;private boolean[][] flag1;private boolean mark;private int[] dx= {0,0,1,-1};private int[] dy= {1,-1,0,0};private boolean[][][] dp;class node{int x,y,xx,yy,step;public node(int x,int y,int xx,int yy,int step) {this.x=x;this.y=y;this.xx=xx;this.yy=yy;this.step=step;}}public int minPushBox(char[][] grid) {int m=grid.length;int n=grid[0].length;flag=new boolean[m][n];flag1=new boolean[m][n];dp=new boolean[4][m][n];int st1x=0,enx=0,st1y=0,eny=0,st2x=0,st2y=0;for(int i=0;i<m;i++)for(int j=0;j<n;j++) {if(grid[i][j]=='S') {st1x=i;st1y=j;}else if(grid[i][j]=='B') {st2x=i;st2y=j;}else if(grid[i][j]=='T') {enx=i;eny=j;}if(grid[i][j]=='#') flag[i][j]=true;}Queue<node> q=new LinkedList<>();q.add(new node(st1x,st1y,st2x,st2y,0));while(!q.isEmpty()) {node now=q.poll();if(now.xx==enx && now.yy==eny)return now.step;for(int i=0;i<4;i++) {int xx=now.xx+dx[i];int yy=now.yy+dy[i];if(xx<0 || xx>=m || yy<0 || yy>=n || flag[xx][yy] || dp[i][now.xx][now.yy])continue;mark=false;flag1[now.x][now.y]=true;flag[now.xx][now.yy]=true;if(i==0) {dfs2(grid,now.x,now.y,now.xx,now.yy-1,m,n);flag1[now.x][now.y]=false;flag[now.xx][now.yy]=false;if(mark) q.add(new node(now.xx,now.yy,xx,yy,now.step+1));}else if(i==1) {dfs2(grid,now.x,now.y,now.xx,now.yy+1,m,n);flag1[now.x][now.y]=false;flag[now.xx][now.yy]=false;if(mark) q.add(new node(now.xx,now.yy,xx,yy,now.step+1));}else if(i==2) {dfs2(grid,now.x,now.y,now.xx-1,now.yy,m,n);flag1[now.x][now.y]=false;flag[now.xx][now.yy]=false;if(mark) q.add(new node(now.xx,now.yy,xx,yy,now.step+1));}else{dfs2(grid,now.x,now.y,now.xx+1,now.yy,m,n);flag1[now.x][now.y]=false;flag[now.xx][now.yy]=false;if(mark) q.add(new node(now.xx,now.yy,xx,yy,now.step+1));}if(mark) dp[i][now.xx][now.yy]=true;mark=false;for(int ii=0;ii<m;ii++)for(int j=0;j<n;j++)flag1[ii][j]=false;}}return -1;}private void dfs2(char[][] grid,int nowpx,int nowpy,int nowx,int nowy,int m,int n) {  	if(nowx<0 || nowx>=m || nowy<0 || nowy>=n || mark) return;if(nowpx==nowx && nowpy==nowy) mark=true;  	for(int i=0;i<4;i++) {if(mark) return;int xx=nowpx+dx[i];int yy=nowpy+dy[i];if(xx<0 || xx>=m || yy<0 || yy>=n || flag[xx][yy] || flag1[xx][yy]) continue;flag1[xx][yy]=true;dfs2(grid,xx,yy,nowx,nowy,m,n);}}
}