Key Task

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.

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

1 3
*#X

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

0 0

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

//109MS	556K
#include<stdio.h>
#include<queue>
#include<string.h>
#include<algorithm>
#define M 107
using namespace std;
int n,m,s,t;
int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
char g[M][M];
bool vis[M][M][17];
char up[4]={'B','Y','R','G'};
char low[4]={'b','y','r','g'};
struct node
{
    int step,x,y,key;
};
int bfs()
{
    queue<node>q;
    node now,next;
    now.x=s;now.y=t;now.step=now.key=0;
    vis[s][t][0]=true;
    q.push(now);
    while(!q.empty())
    {
        now=q.front();
        q.pop();
        if(g[now.x][now.y]=='X')return now.step;
        for(int i=0;i<4;i++)
        {
            next.x=now.x+dir[i][0];
            next.y=now.y+dir[i][1];
            next.step=now.step+1;
            next.key=now.key;
            if(next.x<1||next.x>n||next.y<1||next.y>m||g[next.x][next.y]=='#')continue;//假设越界
            if(g[next.x][next.y]>='A'&&g[next.x][next.y]<='Z'&&g[next.x][next.y]!='X')
            {
                for(int j=0;j<4;j++)
                    if(g[next.x][next.y]==up[j])
                    {
                        if(next.key&(1<<j)&&!vis[next.x][next.y][next.key])//假设没有訪问过且拥有此锁的钥匙
                        {
                            vis[next.x][next.y][next.key]=true;
                            q.push(next);
                        }
                        break;
                    }
            }
            else if(g[next.x][next.y]>='a'&&g[next.x][next.y]<='z')
            {
                for(int j=0;j<4;j++)
                    if(g[next.x][next.y]==low[j])
                    {
                        if((next.key&(1<<j))==0)//假设没有此钥匙
                            next.key+=(1<<j);
                        if(!vis[next.x][next.y][next.key])
                        {
                            vis[next.x][next.y][next.key]=true;
                            q.push(next);
                        }
                    }
            }
            else
            {
                if(!vis[next.x][next.y][next.key])
                {
                    vis[next.x][next.y][next.key]=true;
                    q.push(next);
                }
            }
        }
    }
    return -1;
}
int main()
{
    while(scanf("%d%d",&n,&m),n|m)
    {
        memset(vis,false,sizeof(vis));
        for(int i=1;i<=n;i++)
        {
            scanf("%s",g[i]+1);
            for(int j=1;j<=m;j++)
                if(g[i][j]=='*'){s=i;t=j;}
        }
        int ans=bfs();
        if(ans==-1)printf("The poor student is trapped!\n");
        else printf("Escape possible in %d steps.\n",ans);
    }
    return 0;
}

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