【杭电1242】广搜+队列

 Rescue

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit  Status

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 
      

 

#include<iostream>
#include<cstring>
#include<queue>
using namespace std;
int vis[200][200],sx,sy,n,m,prove;
char map[200][200];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int judge(int x,int y)
{
	if(x>=n||x<0||y<0||y>=m)
	return 0;
	else
	return 1;
}
struct node
{
	int x,y; 
	int time;
}s,e;
queue<node>q;
void bfs()
{
	while(!q.empty())
	q.pop();
	memset(vis,0,sizeof(vis));
	s.time=0;s.x=sx;s.y=sy;
	q.push(s);//把所有r的位置存入队列 
	vis[s.x][s.y]=1;//标记为走过的 
	while(!q.empty())
	{
		s=q.front();
		q.pop();
		if(map[s.x][s.y]=='a')//找到Angle,输出时间 
		{
			cout<<s.time<<endl;
			prove=1;
		}
		if(map[s.x][s.y]=='x')//遇到警卫，先打败，再次进入队列 
		{
			map[s.x][s.y]='.';//变为道路 
			s.time+=1;
			q.push(s);
			continue;
		}
		for(int i=0;i<4;i++)
		{
			e.x=s.x+dx[i];
			e.y=s.y+dy[i];
			if(map[e.x][e.y]=='#'||judge(e.x,e.y)==0||vis[e.x][e.y]==1)
			continue;
			//else 
			e.time=s.time+1;
			q.push(e);
			vis[e.x][e.y]=1;
			
		}
	}
}
int main()
{
	while(cin>>n>>m)
	{
		for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
		{
			cin>>map[i][j];
			if(map[i][j]=='r')
			{
				sx=i;sy=j;//记录朋友的位置 ，不止一个 
			}
		}
		prove=0;
		bfs();
		if(prove==0)
		cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
	}
	return 0;
}