寒假训练——搜索 E - Bloxorz I

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
 
题目：Bloxorz I
大意：（非博主翻译）

šBloxorz是一个风靡世界的小游戏。Bloxorz的地图是一个N行M列的矩阵，每个位置可能是硬地（用.表示）、易碎地面（用E表示）、禁地（用#表示）、起点（用X表示）或终点（用O表示）。你的任务是操作一个1*1*2的长方体。这个长方体在地面上有两种放置形式，“立”在地面上（1*1的面接触地面）或者“躺”在地面上（1*2的面接触地面）。在每一步操作中，可以按上下左右四个键之一。按下之后，长方体向对应的方向沿着棱滚动90度。任意时刻，长方体不能有任何部位接触禁地（否则就会掉下去），并且不能立在易碎地面上（否则会因为压强太大掉下去）。X标识长方体的起始位置，地图上可能有一个X或者两个相邻的X。地图上唯一的一个O标识目标位置。求把长方体移动到目标位置（即立在O上）所需要的最少步数。如果无解，输出Impossible。在移动过程中，X和O标识的位置都可以看作是硬地被利用，3<=N,M<=500。

有两个相邻的X，理解为横躺。
思路：
就是一个bfs，不过不太一样，可以这么理解一下，如果X是正方体和平常迷宫问题有区别吗？这个题目只是X是长方体，
所以在id不同的位置，上下左右移动也不同。
X，若只有一个说明是竖放，两个则说明是延x轴或者y轴放。
X会上下左右移动，这时候就麻烦一点了，因为以前的vis数组是一个二维的直接存放位置即可，但是这个时候若是横放，则会占据
两个位置，不过不用急，很简单就可以想到的，用三维数组，以左上角为标准，和id一起构成判断标准，这样就不会重判了。
具体代码：
在main函数里面，先读取这个图像，然后根据X判断出横躺还是竖放，其次读取O的位置，这个为目标位置
在bfs函数里面，先将这个X的位置存放进去，然后进去搜索。
有一个check函数，就是把这个搜索之后的数读进去，判断，有没有到禁格里，有没有超出去，简单来说，就是是否合法。

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std;
int n,m;
bool vis[5][505][505];
char a[505][505];
struct node
{
    int id;
    int r1,c1,r2,c2;
    int step;
    node(int id,int r1,int c1,int r2,int c2,int step):id(id),r1(r1),c1(c1),r2(r2),c2(c2),step(step){}
};
node ed=node(0,0,0,0,0,0),exa=node(0,0,0,0,0,0);
int to[3][4][4]={
                 {{-2,0,-1,0},{1,0,2,0},{0,1,0,2},{0,-2,0,-1}},
                 {{-1,0,-1,0},{1,0,1,0},{0,2,0,1},{0,-1,0,-2}},
                 {{-1,0,-2,0},{2,0,1,0},{0,1,0,1},{0,-1,0,-1}}
                 };
queue<node>que;
int cmp(int &x1,int &y1,int &x2,int &y2)//确定是竖放还是横躺
{
    if(x1==x2&&y1==y2) return 1;
    else if(x1==x2)
    {
        if(y1>y2) swap(y1,y2);
        return 2;
    }
    else
    {
        if(x1>x2) swap(x1,x2);
        return 3;
    }
}
void check()//检查是否合法
{
    if(a[exa.r1][exa.c1]=='#') return ;
    if(a[exa.r2][exa.c2]=='#') return;
    if(exa.id==1&&(a[exa.r1][exa.c1]=='E'||a[exa.r2][exa.c2]=='E')) return ;
    if(vis[exa.id][exa.r1][exa.c1]) return ;
    if(exa.r1<1||exa.r2<1||exa.r1>n||exa.r2>n||exa.c1<1||exa.c2<1||exa.c1>m||exa.c2>m) return ;
    vis[exa.id][exa.r1][exa.c1]=1;
    que.push(exa);
}
int bfs()//搜索，和普通一样的写法
{
    while(!que.empty()) que.pop();
    memset(vis,0,sizeof(vis));
    vis[ed.id][ed.r1][ed.c1]=1;
    que.push(ed);
    while(!que.empty())
    {
        ed=que.front();
        //printf("%d (%d,%d),(%d,%d)\n",ed.id,ed.r1,ed.c1,ed.r2,ed.c2);
        que.pop();
        if(ed.id==1&&a[ed.r1][ed.c2]=='O') return ed.step;
        for(int i=0;i<4;i++)
        {
            exa.r1=ed.r1+to[ed.id-1][i][0];
            exa.c1=ed.c1+to[ed.id-1][i][1];
            exa.r2=ed.r2+to[ed.id-1][i][2];
            exa.c2=ed.c2+to[ed.id-1][i][3];
            exa.step=ed.step+1;
            exa.id=cmp(exa.r1,exa.c1,exa.r2,exa.c2);
 
            check();
           // printf("id=%d (%d,%d),(%d,%d) i=%d\n",exa.id,exa.r1,exa.c1,exa.r2,exa.c2,i);
        }
    }
    return -1;
}
 
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF&&n+m)
    {
        int cnt=0,r1,c1,r2,c2;
        for(int i=1;i<=n;i++)
        {
            scanf("%s",a[i]+1);
            for(int j=1;j<=m;j++)
            {
                if(a[i][j]=='X')
                {
                    cnt++;
                    if(cnt==1)
                    {
                        r1=r2=i;
                        c1=c2=j;
                    }
                    else
                    {
                        r2=i;
                        c2=j;
                    }
                }
            }
        }
        int t=cmp(r1,c1,r2,c2);
        ed=node(t,r1,c1,r2,c2,0);
        int exe=bfs();
        if(exe==-1) printf("Impossible\n");
        else printf("%d\n",exe);
    }
    return 0;
}