hdu ——1242——Rescue

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

 

#include<iostream>#include<queue>using namespace std;#define max 201int n,m,f,res,ans[max][max];char map[max][max];int dir[4][2]={1,0,0,1,-1,0,0,-1};struct node{ int x,y; int time;};void bfs(int xr,int yr){   queue<node> Q; node a,b; a.x=xr; a.y=yr; a.time=0; Q.push(a); while(!Q.empty()) {  int i,xx,yy;  b=Q.front();  Q.pop();  for(i=0;i<4;i++)  {     xx=b.x+dir[i][0];     yy=b.y+dir[i][1];     a.x=xx;     a.y=yy;     if(a.x>0 && a.x<=n && a.y>0 && a.y<=m &&map[a.x][a.y]!='#')     {      if(map[a.x][a.y]=='.')      {       a.time=b.time+1;      }      else if(map[a.x][a.y]=='x')      {       a.time=b.time+2;      }      if(a.time<ans[a.x][a.y])    //判断当前到该点的时间是否小于先前搜索到该点的时间，如果更下则进队列，否则就不进      {       ans[a.x][a.y]=a.time;       Q.push(a);      }      if(map[a.x][a.y]=='r'&& b.time+1<res )//遇到一次能救天使的人 判断当前的时间是否小于先前找到能救她的人的时间少      {       f=1;                               //并且保存小的时间       res=b.time+1;      }     }  } }}int main(){ int i,j,x,y;// freopen("E:\\test.txt","r",stdin); while(cin>>n>>m) {  for(i=1;i<=n;i++)  {   for(j=1;j<=m;j++)   {    cin>>map[i][j];    ans[i][j]=999999;          //初始化每个搜索到每个点的时间为最大值    if(map[i][j]=='a')    {       x=i;       y=j;    }   }  }  f=0;  res=999999;  bfs(x,y);  if(f==0)        // f用于标记是否可以找到一个能救天使的朋友  {   cout<<"Poor ANGEL has to stay in the prison all his life." <<endl;  }  else  {   cout<<res<<endl;  } } return 0;}