Key Task

The Czech Technical University is rather old — you already know that it celebrates 300 years of its existence in 2007. Some of the university buildings are old as well. And the navigation in old buildings can sometimes be a little bit tricky, because of strange long corridors that fork and join at absolutely unexpected places. 

The result is that some first-graders have often di?culties finding the right way to their classes. Therefore, the Student Union has developed a computer game to help the students to practice their orientation skills. The goal of the game is to find the way out of a labyrinth. Your task is to write a verification software that solves this game. 

The labyrinth is a 2-dimensional grid of squares, each square is either free or filled with a wall. Some of the free squares may contain doors or keys. There are four di?erent types of keys and doors: blue, yellow, red, and green. Each key can open only doors of the same color. 

You can move between adjacent free squares vertically or horizontally, diagonal movement is not allowed. You may not go across walls and you cannot leave the labyrinth area. If a square contains a door, you may go there only if you have stepped on a square with an appropriate key before.

Input

The input consists of several maps. Each map begins with a line containing two integer numbers R and C (1 ≤ R, C ≤ 100) specifying the map size. Then there are R lines each containing C characters. Each character is one of the following: 

Note that it is allowed to have 

• more than one exit,
  
• no exit at all,
  
• more doors and/or keys of the same color, and
  
• keys without corresponding doors and vice versa.
  
  
   
• You may assume that the marker of your position (“*”) will appear exactly once in every map. 
  
  There is one blank line after each map. The input is terminated by two zeros in place of the map size.

Output

For each map, print one line containing the sentence “Escape possible in S steps.”, where S is the smallest possible number of step to reach any of the exits. If no exit can be reached, output the string “The poor student is trapped!” instead. 

One step is defined as a movement between two adjacent cells. Grabbing a key or unlocking a door does not count as a step.

Sample Input

1 10
*........X

1 3
*#X

3 20
####################
#XY.gBr.*.Rb.G.GG.y#
####################

0 0

Sample Output

Escape possible in 9 steps.
The poor student is trapped!
Escape possible in 45 steps.

开始以为一把钥匙开完一个门后就不能用了，然后注意到最后一组样例T_T。

所以这个vis数组要用三维的，他表示的是每个点的拥有钥匙的情况。

这里的方法好玄学，用二进制的位表示是否拥有一把钥匙^_^

个人感觉有点像hash去重

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<algorithm>
#include<iostream>
using namespace std;
const int maxn=1e2+5;
char a[maxn][maxn];
int vis[maxn][maxn][32];//在一个点对应的钥匙有几把
int sx,sy;
int n,m;
inline int to(char c)
{

    if(c=='b'||c=='B') return 0;
    if(c=='Y'||c=='y') return 1;
    if(c=='R'||c=='r') return 2;
    if(c=='g'||c=='G') return 3;
}
inline bool check(int x,int y)
{
    if(a[x][y]=='#') return false;
    if(x<0||y<0||x>=n||y>=m) return false;
    return true;
}
int dx[5]={0,0,-1,1};
int dy[5]={1,-1,0,0};
struct node
{
    int x,y;
    int st;
    int k;//点的去重标记
};
void bfs()
{
    node t;
    t.x=sx;
    t.y=sy;
    t.st=0;
    t.k=0;
    a[sx][sy]='.';
    memset(vis,0,sizeof vis);
    queue<node> q;
    vis[sx][sy][t.k]=1;
    q.push(t);
    while(!q.empty())
    {

        node f=q.front();
        //cout<<f.x<<' '<<f.y<<' '<<a[f.x][f.y]<<endl;
        q.pop();
        for(int i=0;i<4;++i)
        {
            int xx=f.x+dx[i];
            int yy=f.y+dy[i];
            if(check(xx,yy))//没有出界
            {
                if(a[xx][yy]=='.'&&!vis[xx][yy][f.k])
                {
                    //cout<<yy<<endl;
                    vis[xx][yy][f.k]=1;
                    node temp;
                    temp.x=xx;
                    temp.y=yy;
                    temp.st=f.st+1;
                    temp.k=f.k;
                    q.push(temp);
                }
                else if(a[xx][yy]>='a'&&a[xx][yy]<='z')//判断有没有钥匙
                {
                    int hh=1<<to(a[xx][yy]);
                    int tt=f.k;
                    if((f.k&hh)==0) //没有这把钥匙,加上这把钥匙
                        tt+=hh;
                    //cout<<f.k<<endl;
                    if(!vis[xx][yy][tt])
                    {
                        //cout<<yy<<endl;
                        vis[xx][yy][tt]=1;
                        node temp;
                        temp.x=xx;
                        temp.y=yy;
                        temp.st=f.st+1;
                        temp.k=tt;
                        q.push(temp);
                    }
                }
                else if(a[xx][yy]>='A'&&a[xx][yy]<='Z')
                {

                    if(a[xx][yy]=='X')
                    {
                        printf("Escape possible in %d steps.\n",f.st+1);
                        return;
                    }
                    int hh=1<<to(a[xx][yy]);

                    if((hh&f.k)&&!vis[xx][yy][f.k])//有这把钥匙并且没来过
                    {
                        vis[xx][yy][f.k]=1;
                        //cout<<xx<<' '<<yy<<endl;
                        node temp;
                        temp.x=xx;
                        temp.y=yy;
                        temp.st=f.st+1;
                        temp.k=f.k;
                        q.push(temp);
                    }
                }
            }
        }
    }
    puts("The poor student is trapped!");
    return ;
}
int main()
{
    while(~scanf("%d%d",&n,&m)&&(n||m))
    {
        for(int i=0;i<n;++i)
            scanf("%s",a[i]);
        for(int i=0;i<n;++i)
            for(int j=0;j<m;++j)
            if(a[i][j]=='*')
        {
            sx=i;
            sy=j;
            break;
        }
        bfs();

    }
    return 0;
}