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.)

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.

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."

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

13

CHEN, Xue

题意：

一个人想逃出去监狱，x是警卫，a是出发点，r是出去的点 。#是强。黑点是可走的点。

碰到警卫要打一秒，每走一步一秒，问最少多少秒逃出去。

思路：优先队列+bfs。。

解析就不写了，都是老套路。

#include <stdio.h>
#include <algorithm>
using namespace std;
#include <string.h>
#include <queue>
#include <stack>
#include <math.h>
char map[202][202];
int vis[202][202];
int n,m,x2,y2;
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
struct node
{
	int x,y;
	int step;
	bool friend operator<(node a,node b)
	{
		return a.step>b.step;
	}
}
;
bool check(int x,int y)
{
	if(x<0 ||x>=n||y<0||y>=m ||vis[x][y]||map[x][y]=='#')
		return 0;
	return 1;
}
int bfs(int x,int y)
{
	int i;
	priority_queue<node>q;
    node st,ed;
    st.x=x;
	st.y=y;
	st.step=0;
	q.push(st);
	while(!q.empty())
	{
		st=q.top();
		q.pop();
		if(st.x==x2&&st.y==y2)
			return st.step;
		for(i=0;i<4;i++)
		{
			ed.x=st.x+dir[i][0];
			ed.y=st.y+dir[i][1];
			if(!check(ed.x,ed.y))
				continue;
			if(map[ed.x][ed.y]!='x')
				ed.step=st.step+1;
			else
				ed.step=st.step+2;
			vis[ed.x][ed.y]=1;
			q.push(ed);
		}
	}
	return 0;
}  
int main()
{
	int i,j,x1,y1;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;i++)
			scanf("%s",map[i]);
		memset(vis,0,sizeof(vis));
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='a')
				{
					x1=i;
					y1=j;
				}
				else if(map[i][j]=='r')
				{
					x2=i;
					y2=j;
				}
			}
			vis[0][0]=1;
			int ans=bfs(x1,y1);
			if(ans)
				printf("%d\n",ans);
			else
				printf("Poor ANGEL has to stay in the prison all his life.\n");
	}
	return 0;
}