poj3322 Bloxorz I

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O - Bloxorz I

Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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Description

Little Tom loves playing games. One day he downloads a little computer game called 'Bloxorz' which makes him excited. It's a game about rolling a box to a specific position on a special plane. Precisely, the plane, which is composed of several unit cells, is a rectangle shaped area. And the box, consisting of two perfectly aligned unit cube, may either lies down and occupies two neighbouring cells or stands up and occupies one single cell. One may move the box by picking one of the four edges of the box on the ground and rolling the box 90 degrees around that edge, which is counted as one move. There are three kinds of cells, rigid cells, easily broken cells and empty cells. A rigid cell can support full weight of the box, so it can be either one of the two cells that the box lies on or the cell that the box fully stands on. A easily broken cells can only support half the weight of the box, so it cannot be the only cell that the box stands on. An empty cell cannot support anything, so there cannot be any part of the box on that cell. The target of the game is to roll the box standing onto the only target cell on the plane with minimum moves.

The box stands on a single cell

The box lies on two neighbouring cells, horizontally

The box lies on two neighbouring cells, vertically

After Little Tom passes several stages of the game, he finds it much harder than he expected. So he turns to your help.

Input

Input contains multiple test cases. Each test case is one single stage of the game. It starts with two integers R and C(3 ≤ R, C ≤ 500) which stands for number of rows and columns of the plane. That follows the plane, which contains R lines and C characters for each line, with 'O' (Oh) for target cell, 'X' for initial position of the box, '.' for a rigid cell, '#' for a empty cell and 'E' for a easily broken cell. A test cases starts with two zeros ends the input.

It guarantees that

• There's only one 'O' in a plane.
• There's either one 'X' or neighbouring two 'X's in a plane.
• The first(and last) row(and column) must be '#'(empty cell).
• Cells covered by 'O' and 'X' are all rigid cells.

Output

For each test cases output one line with the minimum number of moves or "Impossible" (without quote) when there's no way to achieve the target cell.

Sample Input

7 7
#######
#..X###
#..##O#
#....E#
#....E#
#.....#
#######
0 0

Sample Output

10

很水的bfs，要注意，走时候箱子，是沿着最底下四个边，向四个方向进行翻转，这样就可以用三个状态来标记，有横树和站着三种，这样开一个三维数组就足以保存，但这样的速度是不今人满意的，下次写个双向bfs

#include <iostream>
#include <stdio.h>
#include<string.h>
#include <queue>
using namespace std ;
int n,m,map[605][605],endx,endy,visit[600][600][3];

struct boxtree{
    int step,x[2],y[2],state;
};
int fmin(int a,int b)
{
    if(a<b)
    return a;
    else
    return b;
}
boxtree p,temp,start;
int dir[3][4][5]={
    {
		0,-2,0,-1,1,
		1,0,2,0,2,
		0,1,0,2,1,
	    -2,0,-1,0,2
    },
    {
        0,-1,-3,-3,0,
        1,0,1,1,1,
        0,2,-3,-3,0,
        -1,0,-1,1,1
    },
    {
        0,-1,1,-1,2,
        2,0,-3,-3,0,
        0,1,1,1,2,
        -1,0,-3,-3,0,
    }
};
void init()
{
    char str[200];
    int i,j;
    char c;

    start.x[0]=start.x[1]=start.y[0]=start.y[1]=-1;
    gets(str);//吸入回车
    for( i=0;i<n;i++)
    {
        for( j=0;j<m;j++)
        {
            c=getchar();
            if(c=='#')
            {
              map[i][j]=0;//不可放
            }
            else if(c=='.')
            {
                map[i][j]=2;//可以完全放
            }
            else if(c=='E')//只能放一半
            {
                map[i][j]=1;
            }
            else if(c=='O')
            {
                endx=i;
                endy=j;
				map[i][j]=2;//题目起始和终点都是牢固的
            }
            else if( c=='X')
            {
				map[i][j]=2;
                if(start.x[0]==-1)
                {
                    start.x[0]=i;
                    start.y[0]=j;
                    start.state=0; //站着的壮态
                }
                else
                {
                    start.x[1]=i;
                    start.y[1]=j;
                    start.state=1;
                    if((start.x[0]==start.x[1])&&(start.y[0]!=start.y[1]))
                    {
                       start.state=1;//横着的状态
                       start.y[0]=fmin(start.y[0],start.y[1]);//横着的状态选左边为参照点
                    }
                    else if((start.x[0]!=start.x[1])&&(start.y[0]==start.y[1]))
                    {
                        start.state=2;//树着的状态
                        start.x[0]=fmin(start.x[0],start.x[1]);//树着的选上边为参照点
                    }
                }

            }
        }
        gets(str);

    }
	//gets(str);
//	printf("%d%d end\n",endx,endy);
}
bool changestate(int i)
{
    p.x[0]=temp.x[0]+dir[temp.state][i][0];//都是和相应的参照点进行变换
    p.y[0]=temp.y[0]+dir[temp.state][i][1];
    p.x[1]=temp.x[0]+dir[temp.state][i][2];
    p.y[1]=temp.y[0]+dir[temp.state][i][3];
    p.state=dir[temp.state][i][4];

    if(p.state==0)//如果是站的状态
    {
        if((p.x[0]>=0)&&(p.x[0]<n)&&(p.y[0]>=0&&p.y[0]<m)&&(map[p.x[0]][p.y[0]]==2)&&(visit[p.x[0]][p.y[0]][0]==0))
        {
            visit[p.x[0]][p.y[0]][0]=1;//标记为已访问
            return true;
        }
        else
        {
            return false ;
        }
    }
    else if(p.state==1)//如果是横着的状态
    {
        if((p.x[0]>=0)&&(p.x[0]<n)&&(p.y[0]>=0&&p.y[0]<m-1)&&(map[p.x[0]][p.y[0]]!=0)&&(map[p.x[1]][p.y[1]]!=0)&&(visit[p.x[0]][p.y[0]][1]==0))
        {
            visit[p.x[0]][p.y[0]][1]=1;//标记为已访问
            return true;
        }
        else
        {
            return false ;
        }

    }
    else if( p.state==2)//如果是树的状态
    {
        if((p.x[0]>=0)&&(p.x[0]<n-1)&&(p.y[0]>=0&&p.y[0]<m)&&(map[p.x[0]][p.y[0]]!=0)&&map[p.x[1]][p.y[1]]!=0&&(visit[p.x[0]][p.y[0]][2]==0))
        {
            visit[p.x[0]][p.y[0]][2]=1;//标记为已访问
            return true;
        }
        else
        {
            return false ;
        }

        }

    return false ;
}
int  bfs()
{
    int i;
    queue<boxtree> q;
    memset(visit,0,sizeof(visit));

    while(!q.empty())//清空
    {
        q.pop();
    }

    start.step=0;
    q.push(start);
    while(!q.empty())
    {
        temp=q.front();
        q.pop();

        for( i=0;i<4;i++)//4 个方向进行搜
        {

            if(changestate(i))
            {
                p.step=temp.step+1;
                q.push(p);
			//	printf("%d %d %d\n",p.x[0],p.y[0],p.state);
				if((p.x[0]==endx)&&(p.y[0]==endy)&&(p.state==0))
				{
				  printf("%d\n",p.step);
				  return 1;
				}
            }
        }
    }
    printf("Impossible\n");
    return -1;
}
int main()
{

    while(scanf("%d%d",&n,&m)!=EOF&&n&&m)
    {
         init();
         bfs();
    }

    return 0;
}

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