Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 20938    Accepted Submission(s): 7486

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

7 8

#.#####.

#.a#..r.

#..#x...

..#..#.#

#...##..

.#......

........

 

 

Sample Output

13

题意：问最小花时间从起点到终点，其中要到守卫的地方要多花1个时间。

思路：由于其中有若干步要花费2个时间，可以用优先队列存储，走花费时间少的。用DFS枚举的会超时。

收获：优先队列定义结构体优先级。

1 #include
       
         2 #include
        
          3 #include
         
           4 #include
          
            5 #include
           
             6 using namespace std; 7 #define maxn 200+10 8 int n,m,flag; 9 struct Node10 {11     int x,y,cost;12     bool operator < (const Node &a)const//定义时间花费少的优先级高。13     {14         return cost>a.cost;15     }16 };17 Node st,et,f,k;18 int dx[]={
     0,0,-1,1};19 int dy[]={
     1,-1,0,0};20 int vis[maxn][maxn];21 char map[maxn][maxn];22 void bfs()23 {24     priority_queue
            
             q;25 q.push(st);26 while(!q.empty())27 {28 f=q.top();29 for(int i=0;i<4;i++)30 {31 k.x=f.x+dx[i];32 k.y=f.y+dy[i];33 if(!vis[k.x][k.y]&&k.x>=0&&k.x
             
              =0&&k.y