N - Pushing Boxes

Time Limit:2000MS     Memory Limit:131072KB     64bit IO Format:%I64d & %I64u

Submit  Status

Description

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? 

Input

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. 

Output

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). If there is still more than one such sequence, any one is acceptable.

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. 

Sample Input

1 7

SB....T

1 7

SB..#.T

7 11

###########

#T##......#

#.#.#..####

#....B....#

#.######..#

#.....S...#

###########

8 4

....

.##.

.#..

.#..

.#.B

.##S

....

###T

0 0

Sample Output

Maze #1

EEEEE

Maze #2

Impossible.

Maze #3

eennwwWWWWeeeeeesswwwwwwwnNN

Maze #4

swwwnnnnnneeesssSSS
很经典的双重搜索，要注意，当箱子的步数相等时，人走的步数要越少
#include<iostream>

#include<stdio.h>

#include<string>

#include<string.h>

using namespace std;

int n,bp,m,bx,by,add,maxbox,maxstep,allstep,ret,gx,gy,rx,ry,map[25][25],boxdir[4][2]={{0,-1},{1,0},{0,1},{-1,0}},visit[25][25],peoplevisit[25][25];

struct tree2{int i,bp,add;}setqueue[4];

struct tree{int pre,step,allstep,bx,by,rx,ry;string re;}boxq[60000],peopleq[60000];

char peoplewesn(int i)

{

    if(i==0)

    return 'w';

    else if(i==1)

    return 's';

    else if(i==2)

    return 'e';

    else if(i==3)

    return 'n';

    return 'a';

}

int  peoplebfs(int sx,int sy,int ex,int ey)

{

    memset(peoplevisit,0,sizeof(peoplevisit));

    int t,w,x,y,i,xx,yy;

    t=w=1;

    peopleq[1].step=0;

    peopleq[1].pre=-1;

    peopleq[1].allstep=0;

    peopleq[1].bx=ex;//在peopleq中用bx by存人要走的地方

    peopleq[1].by=ey;

    peopleq[1].rx=sx;

    peopleq[1].ry=sy;

    peopleq[1].re="";

	peoplevisit[sx][sy]=1;

    if(sx==ex&&sy==ey)

    {

        bp=1;

		add=2;

        return 2;

    }

    while(t<=w)

    {

        x=peopleq[t].rx;

        y=peopleq[t].ry;

        if(x==ex&&y==ey)

        {

            bp=t;

			//printf("true");

			add=1;

            return 1;

        }

        for(i=0;i<4;i++)

        {

            xx=x+boxdir[i][0];//人和箱子一个方向

            yy=y+boxdir[i][1];

            if((!peoplevisit[xx][yy])&&(map[xx][yy]==0)&&(xx>=1)&&xx<=n&&yy>=1&&yy<=m)

            {

			//	printf("%d i\n",i);

                peoplevisit[xx][yy]=1;//标记为访问

                peopleq[++w].pre=t;

                peopleq[w].rx=xx;

                peopleq[w].ry=yy;

                peopleq[w].bx=ex;

                peopleq[w].by=ey;

                peopleq[w].re=peopleq[t].re+peoplewesn(i);

                peopleq[w].step=peopleq[t].step+1;

            }

        }

        t++;

	//	printf("%d w%d\n",t,w);

    }

//	printf("%d %d false\n",t,w);

	add=0;

    return 0;

}

bool boxcan(int x,int y,int i,int tt)//判断是否能走，并搜索人

{

    int xx,yy;

    xx=x+boxdir[i][0];

    yy=y+boxdir[i][1];

    map[x][y]=1;

    if((visit[xx][yy]==0)&&(map[xx][yy]==0)&&xx>=1&&xx<=n&&yy>=1&&yy<=m&&peoplebfs(boxq[tt].rx,boxq[tt].ry,x-boxdir[i][0],y-boxdir[i][1]))

    {

	//	printf("%d  %d tt%d",tt,boxq[tt].rx,boxq[tt].ry);

        visit[xx][yy]=1;//标记为已访问

        map[x][y]=0;

        return true;

    }

    map[x][y]=0;

    return false ;

}

char boxwesn(int i)//判断箱子走的方向

{

    if(i==0)

    return 'W';

    else if(i==1)

    return 'S';

    else if(i==2)

    return 'E';

    else if(i==3)

    return 'N';

    return 'a';

}

int boxbfs(int sx,int sy,int ex,int ey)//找箱子的最短路

{

	memset(visit,0,sizeof(visit));

    int t,w,x,y,i;

    t=w=1;

    boxq[1].pre=-1;

    boxq[1].step=0;

    boxq[1].bx=sx;

    boxq[1].by=sy;

	visit[sx][sy]=1;

    boxq[1].re="";

    boxq[1].rx=rx;

    boxq[1].ry=ry;

    boxq[1].allstep=0;

    maxbox=-1;

    if(sx==ex&&sy==ey)

    {

		return 2;

    }

    while(t<=w)

    {

        x=boxq[t].bx;

        y=boxq[t].by;

		if(ret==2&&boxq[t].step>maxstep)

		{

			return maxbox;

		}

        if(x==ex&&y==ey)//找到结果，但这里不能停，要把所有的是最小步数的搜出来，并把最小人走的数输出来

        {

			if(ret==1)

			{

				ret=2;

				maxstep=boxq[t].step;

				allstep=boxq[t].allstep;

				maxbox=t;

			}

			if(ret==2&&boxq[t].allstep<allstep)

			{

				maxbox=t;

				allstep=boxq[t].allstep;

			}

         }

		int j;

        for(i=0,j=0;i<4;i++)//４个方向搜

        {

            if(boxcan(x,y,i,t))

            {

				{

				 w++;

                boxq[w].pre=t;

                boxq[w].bx=x+boxdir[i][0];

                boxq[w].by=y+boxdir[i][1];

                boxq[w].rx=x;

                boxq[w].ry=y;

                boxq[w].step=boxq[t].step+1;

                boxq[w].allstep=boxq[t].allstep+peopleq[bp].step;

				if(add==2)//判定是否跟着箱子走

                boxq[w].re=boxq[t].re+boxwesn(i);////结果保存下来

				else if(add==1)

				boxq[w].re=boxq[t].re+peopleq[bp].re+boxwesn(i);

				}

			}

		}

        t++;

    }

	if(ret==2)//如果有了最小值要标记下来

		return 1;

    return -1;//不可达到

}

int main ()

{

    int i,j,tcase=1;

    char c,str[200];

    while(scanf("%d%d",&n,&m)!=EOF)

    {

		getchar();

        ret=1;maxbox=-1;maxstep=-1;allstep=-1;

        if(n==0&&m==0)

        break;

        for(i=1;i<=n;i++)

        {

            for(j=1;j<=m;j++)

            {

                c=getchar();

                if(c=='#')

                    map[i][j]=1;

                else if(c=='.')

                    map[i][j]=0;

                else if(c=='B')

                {

                    map[i][j]=0;

                    bx=i;

                    by=j;

                }

                else if(c=='T')

                {

                    map[i][j]=0;

                    gx=i;

                    gy=j;

                }

                else if(c=='S')

                {

                    map[i][j]=0;

                    rx=i;

                    ry=j;

                }

            }

            gets(str);

        }

        printf("Maze #%d\n",tcase++);

        if(boxbfs(bx,by,gx,gy)==-1)

        {

            printf("Impossible.\n\n");

        }

        else

        {

                cout<<boxq[maxbox].re<<endl<<endl;

        }

    }

    return 0;

}







												


																

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