hdu 1312 Red and Black

Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4374 Accepted Submission(s): 2845

Problem Description

There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can't move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input

The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

'.' - a black tile
'#' - a red tile
'@' - a man on a black tile(appears exactly once in a data set)

Output

For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).

Sample Input

6 9 ....#. .....# ...... ...... ...... ...... ...... #@...# .#..#. 11 9 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#..@#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### ...@... ###.### ..#.#.. ..#.#.. 0 0

Sample Output

45 59 6 13

Source

Asia 2004, Ehime (Japan), Japan Domestic

Recommend

Eddy

 

 

import java.util.Scanner;

public class Main {
 static int px[]={0,1,0,-1};
 static int py[]={1,0,-1,0};
 static int t;
 static public void DFS(int i,int j,char a[][],boolean b[][],int m,int n){
  int k,x,y;
  b[i][j]=true;
  t++;
  for(k=0;k<4;k++){
   x=i+px[k];
   y=j+py[k];
   if(x>=0&&x=0&&y     if(a[x][y]!='#'&&b[x][y]==false){
     DFS(x,y,a,b,m,n);
    }
   }
   
  }
  
 }

 public static void main(String[] args) {
  int m,n,i,j,x1=0,x2=0;
  Scanner oo=new Scanner(System.in);
  while(oo.hasNext()){
   n=oo.nextInt();
   m=oo.nextInt();
   if(m==0&&n==0){break;}
   String s;
   char a[][]=new char[m][n];
   for(i=0;i     s=oo.next();
    for(j=0;j      a[i][j]=s.charAt(j);
     if(a[i][j]=='@'){
      x1=i;x2=j;
     }
    }
   }
   t=0;
      boolean b[][]=new boolean[m][n];
      DFS(x1,x2,a,b,m,n);
      System.out.println(t);
  }
 

 }

}