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.

/*************************************************************************
	> File Name: 3.cpp
	> Author:yuan
	> Mail:
	> Created Time: 2014年11月26日 星期三 13时09分34秒
 ************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<queue>
using namespace std;
char mat[105][105];
int ans;
struct node
{
    int x,y;
    int num,key;
};
bool vis[105][105][30];
queue<node> q;
int sx,sy;
int n,m,top,base;
int next[4][2]={1,0,0,1,-1,0,0,-1};

void BFS()
{
    int x1,x2,y1,y2;node p,qq;
    while(!q.empty()){
     p=q.front();
      x1=p.x;y1=p.y;
        if(mat[x1][y1]=='X'){
            ans=p.num;
            return;
        }
        for(int i=0;i<4;i++)
        {
            int key;
            x2=x1+next[i][0];y2=y1+next[i][1];
            if(x2<0||x2>=n||y2<0||y2>=m||mat[x2][y2]=='#') continue;

            if(mat[x2][y2]=='.'||mat[x2][y2]=='*'||mat[x2][y2]=='X')
            {
                key=p.key;
                if(vis[x2][y2][key]==0)
                {
                    qq.x=x2;qq.y=y2;qq.num=p.num+1;qq.key=p.key;
                    q.push(qq);
                    vis[x2][y2][key]=1;
                }
            }

             else if(mat[x2][y2]=='B'||mat[x2][y2]=='Y'||mat[x2][y2]=='R'||mat[x2][y2]=='G')
              {
                key=p.key;
                if(vis[x2][y2][key]==0){
                    int d;
                    if(mat[x2][y2]=='B') d=0;
                    else  if(mat[x2][y2]=='Y') d=1;
                           else  if(mat[x2][y2]=='R') d=2;
                                  else  if(mat[x2][y2]=='G') d=3;
                    int lock=1<<d;
                    if(lock&key){
                         qq.x=x2;qq.y=y2;qq.num=p.num+1;qq.key=p.key;
                         q.push(qq);
                         vis[x2][y2][key]=1;
                    }
                }
            }
            else if(mat[x2][y2]=='b'||mat[x2][y2]=='y'||mat[x2][y2]=='r'||mat[x2][y2]=='g')
            {
                  int d;
                    if(mat[x2][y2]=='b') d=0;
                    else  if(mat[x2][y2]=='y') d=1;
                           else  if(mat[x2][y2]=='r') d=2;
                                  else  if(mat[x2][y2]=='g') d=3;
                key=1<<d;
                if(vis[x2][y2][p.key]==0){
                    qq.x=x2;qq.y=y2;qq.num=p.num+1;qq.key=p.key|key;
                    q.push(qq);
                    vis[x2][y2][p.key]=1;
                }
            }
        }
      q.pop();
    }
}
int main()
{
     while(1){
        scanf("%d%d",&n,&m);
        if(n==0&&m==0) break;
        memset(mat,0,sizeof(mat));
        memset(vis,0,sizeof(vis));
        while(!q.empty()) q.pop();
        for(int i=0;i<n;i++) scanf("%s",mat[i]);
        for(int i=0;i<n;i++)
          for(int j=0;j<m;j++)
        {
            if(mat[i][j]=='*') {sx=i,sy=j;}

        }
        node p;
        p.x=sx;p.y=sy;p.num=0;p.key=0;
        q.push(p);
        vis[sx][sy][0]=1;
        ans=100000000;
        BFS();
        if(ans<100000000) printf("Escape possible in %d steps.\n",ans);
        else printf("The poor student is trapped!\n");
    }
    return 0;
}

HDU 1885 BFS+状态压缩

原文：http://blog.csdn.net/yuanchang_best/article/details/41541603