hdu 1242 Rescue

Rescue

http://acm.hdu.edu.cn/showproblem.php?pid=1242

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4611    Accepted Submission(s): 1708

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
   
   
   
   

 晕，被水题给水了

代码：

//hdu 1242 Rescue 记住先判断在往各个方向扫描！！！
//还有memset(mark,0,sizeof(mark));
//不要写成memset(mark,0,sizeof(mark[0][0]));!!!!×××

#include<iostream>
#include<queue>
using namespace std;
int n,m;
int i,j;
char map[210][210];
int mark[210][210];
int dir[][2]={0,1,1,0,0,-1,-1,0};
struct node
{
 int x;
 int y;
 int step;
}begin;

void dfs()
{
    int i;
    node p,temp;
    queue<node> q;
    q.push(begin);
    while(q.size())
    {
        p=q.front();
        q.pop();
        if('r'==map[p.x][p.y])
  {
   printf("%d/n",p.step);
   return ;  //退出函数
  }
  
  if('x'==map[p.x][p.y])  //留空退步
  {
   map[p.x][p.y]='.';
   p.step++;
   q.push(p);
  }
  else
   for(i=0;i<4;i++)
   {
    temp.x=p.x+dir[i][0];
    temp.y=p.y+dir[i][1];
    if(temp.x>n||temp.x<0||temp.y>m||temp.y<0||map[temp.x][temp.y]=='#'||mark[temp.x][temp.y]!=0)   //边界判断
     continue;
                temp.step=p.step+1;
                mark[temp.x][temp.y]=1;
                q.push(temp);
            }
 }
 
 printf("Poor ANGEL has to stay in the prison all his life./n");
}
int main()
{
 while(scanf("%d%d",&n,&m)!=EOF)
 {
  memset(mark,0,sizeof(mark));
  
  
  for(i=0;i<n;i++)
  {
   getchar();
   for(j=0;j<m;j++)
   {  
    scanf("%c",&map[i][j]);
    if(map[i][j]=='a') //找出起点
    {
     begin.x=i;
     begin.y=j;
     begin.step=0;
    }
   }
  }
  //  for(i=0;i<n;i++)
  //  {
  //   cout<<map[i]<<endl;
  //  }
  
  dfs(); //dfs
  
  
 }
 return 0;
}