Rescue

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

Sample Input

7 8
#.#####.
#.a#..r.
#..#x...
..#..#.#
#...##..
.#......
........

Sample Output

13

这道题是个搜索，题意大家都能理解，但是需要注意的是你得杀死守卫，才能拯救天使。好了，话不多说上代码，

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
using namespace std;
char s[1010][1010];// 地图的录入数组
int n,m,startx,starty,endx,endy;//起始点和终点的坐标
int book[1010][1010],a[4][2]= {0,1,1,0,0,-1,-1,0};//标记走过的位置，以及方向数组
struct node
{
    int x,y,step;
};
void cn()
{
    queue<node>Q;
    node st,en;
    st.x=startx;
    st.y=starty;
    st.step=0;
    Q.push(st);
    while(!Q.empty())
    {
        en=Q.front();
        Q.pop();
        for(int i=0; i<4; i++)
        {
            int tx=en.x+a[i][0];
            int ty=en.y+a[i][1];
            if(tx<0||ty<0||tx>=n||ty>=m||s[tx][ty]=='#')//碰到边界或者墙
                continue;
            st.x=tx;
            st.y=ty;
            st.step=en.step+1;
            if(s[tx][ty]=='x')//杀死守卫
                st.step++;
            if(!book[tx][ty]||st.step<book[tx][ty])//这步是关键，利用标记数组，既记录了走过的点，还记录了花费的步数。
            {
                book[tx][ty]=st.step;
                Q.push(st);
            }
        }
    }
    if(!book[endx][endy])
        printf("Poor ANGEL has to stay in the prison all his life.\n");
    else
        printf("%d\n",book[endx][endy]);
    return;
}
int main()
{
    int i,j,x1,y1;
    while(~scanf("%d%d",&n,&m))
    {
         memset(book,0,sizeof(book));
        for(i=0; i<n; i++)
        {
            scanf("%s",s[i]);
            for(j=0; j<m; j++)
            {
                if(s[i][j]=='r')
                {
                    startx=i;
                    starty=j;
                }
                else if(s[i][j]=='a')
                {
                    endx=i;
                    endy=j;
                }
            }
        }
        book[startx][starty]=-1;//起始点的距离相当于负数
        cn();
    }
    return 0;
}