描述
Imagine you are standing inside a two-dimensional maze composed of square cells which may or may not be filled with rock. You can move north, south, east or west one cell at a step. These moves are called walks.
One of the empty cells contains a box which can be moved to an adjacent free cell by standing next to the box and then moving in the direction of the box. Such a move is called a push. The box cannot be moved in any other way than by pushing, which means that if you push it into a corner you can never get it out of the corner again.

One of the empty cells is marked as the target cell. Your job is to bring the box to the target cell by a sequence of walks and pushes. As the box is very heavy, you would like to minimize the number of pushes. Can you write a program that will work out the best such sequence?

輸入
The input contains the descriptions of several mazes. Each maze description starts with a line containing two integers r and c (both <= 20) representing the number of rows and columns of the maze.

Following this are r lines each containing c characters. Each character describes one cell of the maze. A cell full of rock is indicated by a ‘#‘ and an empty cell is represented by a ‘.‘. Your starting position is symbolized by `S‘, the starting position of the box by ‘B‘ and the target cell by ‘T‘.

Input is terminated by two zeroes for r and c.
輸出
For each maze in the input, first print the number of the maze, as shown in the sample output. Then, if it is impossible to bring the box to the target cell, print ``Impossible.‘‘.

Otherwise, output a sequence that minimizes the number of pushes. If there is more than one such sequence, choose the one that minimizes the number of total moves (walks and pushes). 本題沒有 Special Judge，多解時，先最小化箱子被推動的次數，再最小化人移動的步數。若仍有多條路線，則按照N、S、W、E的順序優先選擇箱子的移動方向（即先上下推，再左右推）。在此前提下，再按照n、s、w、e的順序優先選擇人的移動方向（即先上下動，再左右動）。

Print the sequence as a string of the characters N, S, E, W, n, s, e and w where uppercase letters stand for pushes, lowercase letters stand for walks and the different letters stand for the directions north, south, east and west.

Output a single blank line after each test case.
樣例輸入

1 7
SB....T
1 7
SB..#.T
7 11
###########
#T##......#
#.#.#..####
#....B....#
#.######..#
#.....S...#
###########
8 4
....
.##.
.#..
.#..
.#.B
.##S
....
###T
0 0

樣例輸出

Maze #1
EEEEE

Maze #2
Impossible.

Maze #3
eennwwWWWWeeeeeesswwwwwwwnNN

Maze #4
swwwnnnnnneeesssSSS

來源
Southwestern European Regional Contest 1997
本題數據為加強版，在原比賽或POJ AC的程序有可能只得50分

題解：
什麽是bfs套bfs呢？在這題裏，把人和物分開來想，用物體位置+人的方向來存儲人與物之間的關系，那麽人從左推物體就相當於人先走到物體的左邊，然後把物體退了一下。
對人的走的計算也通過bfs來實現，所以是bfs套bfs。
代碼實現容易出錯，一些我想了比較久的代碼都加了註釋，希望對大家有所幫助。

#include <bits/stdc++.h>
#define int long long
using namespace std;
int n,m,t;
int dx[]={-1,1,0,0},dy[]={0,0,-1,1};
char ch[23][23],A[]={'N','S','W','E'},a[]={'n','s','w','e'};
bool vis[23][23];
string rec;
struct node {
    int x,y,px,py;
    string ans;
};
queue <node> q;
bool ok(int x,int y) {
    return x>0 && x<=n && y>0 && y<=m && ch[x][y]!='#';
}
bool bfs2(node st,node ed) {//東西從pre推到now
    rec="";
    node fir;
    fir.x=st.px,fir.y=st.py;
    fir.ans="";
    while(!q.empty()) q.pop();
    q.push(fir);
    memset(vis,0,sizeof vis);
    while(!q.empty()) {
        node now=q.front(),nxt;
        q.pop();
        if(now.x==st.x && now.y==st.y) {//人在物的位置了（說明物體已經被推過去了）
            rec=now.ans;
            return 1;
        }
        for(int i=0;i<4;i++) {
            nxt=now;
            nxt.x+=dx[i],nxt.y+=dy[i];
            if(!ok(nxt.x,nxt.y) || (nxt.x==ed.x && nxt.y==ed.y) || vis[nxt.x][nxt.y]) continue;//人到物體該去的地方了，顯然不行
            vis[nxt.x][nxt.y]=1;
            nxt.ans=now.ans+a[i];
            q.push(nxt);
        }
    }
    return 0;
}
string solve() {
    node fir;
    fir.ans="";
    fir.x=fir.y=fir.px=fir.py=-1;
    for(int i=1;i<=n && (fir.x==-1 || fir.px==-1);i++)
        for(int j=1;j<=m && (fir.x==-1 || fir.py==-1);j++)
            if(ch[i][j]=='B') fir.x=i,fir.y=j,ch[i][j]='.';
            else if(ch[i][j]=='S') fir.px=i,fir.py=j,ch[i][j]='.';
    queue <node> qq;
    qq.push(fir);
    bool vv[23][23][5];
    memset(vv,0,sizeof vv);
    string ans="Impossible.";
    int cntans=0x3f3f3f3f,cnt=0x3f3f3f3f;
    while(!qq.empty()) {
        node now=qq.front(),nxt,pre;
        qq.pop();
        if(ch[now.x][now.y]=='T') {//更新答案
            int cntnow=0;
            for(int i=0;i<now.ans.length();i++) if(now.ans[i]>='A' && now.ans[i]<='Z') cntnow++;
            if(cntnow<cntans || (cntnow==cntans && now.ans.length()<cnt)) {//題目中的排序方法
                ans=now.ans;
                cntans=cntnow;
                cnt=now.ans.length();
            }
            continue;
        }
        for(int i=0;i<4;i++) {
            nxt=now;
            nxt.x+=dx[i];
            nxt.y+=dy[i];
            if(!ok(nxt.x,nxt.y) || vv[nxt.x][nxt.y][i]) continue;
            pre=now;
            if(i==0) pre.x++;
            else if(i==1) pre.x--;
            else if(i==2) pre.y++;
            else if(i==3) pre.y--;//找到上一步狀態
            if(!bfs2(pre,now)) continue;
            vv[nxt.x][nxt.y][i]=1;
            nxt.ans=now.ans+rec;//人動
            nxt.ans+=A[i];//箱子動
            nxt.px=now.x;
            nxt.py=now.y;
            qq.push(nxt);
        }
    }
    return ans;
}
signed main() {
    while(scanf("%lld%lld",&n,&m)!=EOF && n && m) {
        for(int i=1;i<=n;i++) scanf("%s",ch[i]+1);
        cout << "Maze #" << ++t << '\n' << solve() << '\n' << '\n';
    }
    return 0;
}

POJ1475 Pushing Boxes(BFS套BFS)