Problem Description

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

 

Sample Output

 

Author

CHEN, Xue

 

Source

ZOJ Monthly, October 2003

 

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救天使的最短时间。

陷阱：天使的朋友可能不止一个，所以如果以朋友为起点，时间效率大大降低，
所以需要反向遍历。因为我们并不知道朋友的数量如果每个点都要去枚举的话，可能会超时，所以这时候我们就需要用到剪枝操作，我们直接从终点开始遍历，只要遇到朋友的位置就直接返回，这样就会快的很多，这里我们用一个优先队列去操作一番，每次都取出最小的，看看能不能满足要求，一旦满足要求就直接返回。

#include<iostream>
#include<queue>
#include<vector>
using namespace std;
int dir_x[4]={0,0,1,-1};
int dir_y[4]={1,-1,0,0};
class location{public:int x;int y;int steps;friend bool operator <(location l_1,location l_2){return l_1.steps>l_2.steps;}
};
bool limit(int x,int y,int n,int m){if(x<=0||y<=0||x>n||y>m) return 0;else return 1;
}
location BFS(vector<vector<char>>& map,int x_start,int y_start,int n,int m){location l_current,l_next;priority_queue<location> q_1;vector<vector<int>> sign(210,vector<int>(210,1));l_current.x=x_start;l_current.y=y_start;l_current.steps=0;q_1.push(l_current);while(!q_1.empty()){l_current=q_1.top();q_1.pop();sign[l_current.x][l_current.y]=0;if(map[l_current.x][l_current.y]=='r'){return l_current;}for(int i=0;i<4;++i){l_next.x=l_current.x+dir_x[i];l_next.y=l_current.y+dir_y[i];if(sign[l_next.x][l_next.y]&&limit(l_next.x,l_next.y,n,m)&&map[l_next.x][l_next.y]!='#'){sign[l_next.x][l_next.y]=0;   if(map[l_next.x][l_next.y]=='x'){l_next.steps=l_current.steps+2;}else l_next.steps=l_current.steps+1;q_1.push(l_next);}}}l_current.steps=-1;return l_current;
}
int main(){int n,m;while(cin>>n>>m){vector<vector<char>> map(210,vector<char>(210,0));int x_start,y_start,x_end,y_end;for(int i=1;i<=n;++i){for(int j=1;j<=m;++j){cin>>map[i][j];if(map[i][j]=='a'){x_start=i;y_start=j;}}}location res=BFS(map,x_start,y_start,n,m);if(res.steps==-1) cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;else cout<<res.steps<<endl;}return 0;
}