【BFS】营救天使

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.) 

InputFirst 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. 
OutputFor 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JAVA写的队列实现的BFS算法:

注意一下队列的声明方法:

Queue<Integer> queue=new LinkedList<>();
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class Main{
  final static int [][]direction={{-1,0},{1,0},
                  {0,-1},{0,1}};
  static int X,Y;//行、列
  static int startX,startY;
  static char [][]mapp=new char[210][210];
  static boolean [][]visit=new boolean[210][210];
  
  public static int Bfs() {
    Queue<Integer> queue=new LinkedList<>();
      queue.add(startX);
      queue.add(startY);
      queue.add(0);//压入时间
      visit[startX][startY]=true;
      while(!queue.isEmpty())
      {
        int r=queue.poll();
        int c=queue.poll();
        int t=queue.poll();
        //System.out.println(r+" "+c+" "+t);
        if(mapp[r][c]=='a')
          return t;
        else if(mapp[r][c]=='x')//杀死guard
        {
          mapp[r][c]='.';
          queue.add(r);
          queue.add(c);
          queue.add(t+1);
          continue;
        }
        //正式BFS
        for(int d=0;d<4;d++)
        {
          int dx=r+direction[d][0];
          int dy=c+direction[d][1];
          if(dx>=0 && dx<X && dy>=0 && dy<Y)
          {
            if(mapp[dx][dy]!='#' && !visit[dx][dy])
            {
              visit[dx][dy]=true;
              queue.add(dx);
              queue.add(dy);
              queue.add(t+1);
            }
          }
        }
      }
      return -1;
    }
  public static void main(String []args) {
    Scanner cin=new Scanner(System.in);
    while(cin.hasNext())
    {
      X=cin.nextInt();
      Y=cin.nextInt();
      cin.nextLine();//吸收回车
      for(int i=0;i<X;i++)
      {
        mapp[i]=cin.nextLine().toCharArray();
        //System.out.println(mapp[i]);
        for(int k=0;k<Y;++k)
        {
          if(mapp[i][k]=='r'){
            startX=i;
            startY=k;
          }
          
          visit[i][k]=false;
        }
      }
      //System.out.println(startX+" "+startY);
      int time=Bfs();
      if(time==-1)
        System.out.println("Poor ANGEL has to stay in the prison all his life.");
      else System.out.println(time);
    }
    cin.close();
  }
}

 

 

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