Rescue

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

用优先队列写，一开始错了，然后想起来，如果朋友有多个，就不对了，但是天使只有一个，所以可以从天使去找朋友，找到就返回。

#include<bits/stdc++.h>
using namespace std;
int n,m;
char Map[220][220];
int vis[220][220];
int dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}}; //走4个方向 
int ex,ey;//天使的位置 

struct node
{
	int x,y,step;
	friend bool operator<(node a,node b)//优先队列按照时间大小排序
	{
		return a.step>b.step;
	}
};

bool check(int x,int y) //判断能不能走 
{
	if(x>=0&&x<n&&y>=0&&y<m&&vis[x][y]==0&&Map[x][y]!=\'#\')
		return 1;
	return 0;
}

int bfs(int x,int y)
{
	priority_queue<node>q;
	node now,next;
	now.x=x;
	now.y=y;
	now.step=0;
	vis[now.x][now.y]=1;
	q.push(now);
	while(!q.empty())
	{
		now=q.top();
		q.pop();
		if(Map[now.x][now.y]==\'r\') //找到朋友就返回 
		{
			return now.step;
		}
		for(int i=0;i<4;i++)
		{
			next.x=now.x+dir[i][0];
			next.y=now.y+dir[i][1];
			if(check(next.x,next.y))
			{
				if(Map[next.x][next.y]==\'x\') //杀人加2 
				{
					vis[next.x][next.y]=1;
					next.step=now.step+2;
					q.push(next); 
				}
				else
				{
					vis[next.x][next.y]=1;
					next.step=now.step+1;
					q.push(next);
				}
			}
		}
	}
	return -1;
}

int main()
{
	while(scanf(\"%d%d\",&n,&m)!=EOF)
	{
		getchar();
		memset(vis,0,sizeof(vis));
		for(int i=0;i<n;i++)
		{
			for(int j=0;j<m;j++)
			{
				scanf(\"%c\",&Map[i][j]);
				if(Map[i][j]==\'a\')
				{
					ex=i,ey=j;
				}
			}
			getchar();
		}
		int ans=bfs(ex,ey);
		if(ans==-1)
			printf(\"Poor ANGEL has to stay in the prison all his life.\\n\");
		else
		    printf(\"%d\\n\",ans);
	}
	return 0;
}

（宿題はとても多すぎたんです）