BFS题，由于其存在可以消灭警卫的问题，所以其需要求出地图中每个方格的最小步数。更新的时候应该是如果当前值小于 格子的值才进行更新操作，否则是不用更新的。

代码：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
using namespace std;
const int maxn=205;
struct node
{int x;int y;int ans;
};
char map[maxn][maxn];
int r,c,movex[4]={1,-1,0,0},movey[4]={0,0,-1,1},ans[maxn][maxn];
queue<node> q;
int main()
{while(scanf("%d%d",&r,&c)!=EOF){memset(ans,-1,sizeof(ans));for(int i=0;i<=r+1;i++)map[i][0]=map[i][c+1]='#';for(int i=0;i<=c+1;i++)map[0][i]=map[r+1][i]='#';int ex,ey;for(int i=1;i<=r;i++){getchar();for(int j=1;j<=c;j++){scanf("%c",&map[i][j]);if(map[i][j]=='a'){q.push((node){i,j,0});}else if(map[i][j]=='r'){ex=i;ey=j;}}}while(!q.empty()){node ita=q.front();q.pop();for(int i=0;i<4;i++){int x=ita.x+movex[i];int y=ita.y+movey[i];if(map[x][y]=='#')continue;if(ans[x][y]==-1||ans[x][y]>ita.ans+1){if(map[x][y]=='x'){q.push((node){x,y,ita.ans+2});ans[x][y]=ita.ans+2;}else{q.push((node){x,y,ita.ans+1});ans[x][y]=ita.ans+1;}}}}if(ans[ex][ey]==-1)printf("Poor ANGEL has to stay in the prison all his life.\n");elseprintf("%d\n",ans[ex][ey]);}return 0;
}