bfs。 一把某种颜色的锁开 所有这个颜色的门。

状态检查压缩一下  vis[][][2^4];

跟HDU 1429 类似。至于颜色判断我用了 map；

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<algorithm>
#include<map>
#include<stack>
#include<iostream>
#include<list>
#include<set>
#include<vector>
#include<cmath>

#define INF 0x7fffffff
#define eps 1e-8
#define LL long long
#define PI 3.141592654
#define CLR(a,b) memset(a,b,sizeof(a))
#define FOR(i,a,n) for(int i= a;i< n ;i++)
#define FOR0(i,a,b) for(int i=a;i>=b;i--)
#define pb push_back
#define mp make_pair
#define ft first
#define sd second
#define sf scanf
#define pf printf
#define acfun std::ios::sync_with_stdio(false)

#define SIZE 100+1
using namespace std;

int xx[]={0,0,-1,1};
int yy[]={-1,1,0,0};

int n,m;
char g[SIZE][SIZE];
map<char,int>Mykey;

struct lx
{
    int x,y;
    int t;
    int key;
    void init(int xx,int yy,int tt,int kk)
    {
        x=xx,y=yy,t=tt,key=kk;
    }
}start;
int num[]={8,4,2,1};
bool cheack(char c,int key)
{
    //g r y b
    bool unlock[4];
    FOR(i,0,4)
    {
        if(key>=num[i])
        {
            unlock[i]=1;
            key-=num[i];
        }
        else
            unlock[i]=0;
    }
    int tmp=Mykey[c];
    int ans=3;
    while(tmp>1)
    {
        tmp/=2;
        ans--;
    }
    return unlock[ans];
}
void bfs()
{
    bool vis[SIZE][SIZE][16];
    CLR(vis,0);
    vis[start.x][start.y][start.key]=1;
    queue<lx>q;
    q.push(start);
    while(!q.empty())
    {
        lx tmp=q.front();
        q.pop();
        //pf("%d %d time=%d\n",tmp.x,tmp.y,tmp.t);
        if(g[tmp.x][tmp.y]=='X')
        {
            pf("Escape possible in %d steps.\n",tmp.t);
            return;
        }
        FOR(k,0,4)
        {
            int x=tmp.x+xx[k];
            int y=tmp.y+yy[k];
            int key=tmp.key;
            if(x<0||y<0||x>=n||y>=m||g[x][y]=='#')continue;

            if(g[x][y]>='a'&&g[x][y]<='z'&&!cheack(g[x][y],key))
                key+=Mykey[ g[x][y] ];
            else if(g[x][y]>='A'&&g[x][y]<='Z'&&g[x][y]!='X')
            {
                if(!cheack(g[x][y]-'A'+'a',key))continue;
            }
            if(vis[x][y][key])continue;
            lx now;
            now.init(x,y,tmp.t+1,key);
            vis[x][y][key]=1;
            q.push(now);
        }
    }
    puts("The poor student is trapped!");
}

int main()
{
    Mykey['b']=1;
    Mykey['y']=2;
    Mykey['r']=4;
    Mykey['g']=8;
    while(~sf("%d%d",&n,&m),n||m)
    {
        FOR(i,0,n)
        {
            char str[SIZE];
            sf("%s",str);
            FOR(j,0,m)
            {
                g[i][j]=str[j];
                if(str[j]=='*')
                    start.init(i,j,0,0);
            }
        }
        bfs();
    }
}

HDU 1885 Key Task