poj1649 Rescue(BFS+优先队列)

poj1649 Rescue(BFS+优先队列)

Rescue

Time Limit: 2 Seconds       Memory Limit: 65536 KB

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

广度优先搜索找最短时间。

看代码注释吧

#include #include #include using namespace std;struct node{ int x,y,time;//x,y方格的位置。time当前所有的时间 friend bool operator<(node a,node b)//优先队列按照时间大小排序 { return a.time>b.time; }};priority_queues;char map[205][205];//地图int m,n,vis[205][205],escape,dir[4][2]={0,1,0,-1,1,0,-1,0};//vis标记，dir表示四个方向bool judge(int x,int y)//判断当前位置时候可以走{ if(x>=0&&y>=0&&x

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