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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13
import java.util.Scanner;
public class P1312 {
public static int n,m,count;
public static int[][] dir={{1, 0}, {0, -1}, { -1, 0}, {0, 1}};//朝四个方向遍历
public static char[][] map=new char[30][30];//地图
public static void main(String[] args) {
Scanner sc=new Scanner(System.in);
while(sc.hasNext()){
count=0;
m=sc.nextInt();
n=sc.nextInt();
if(m==0&&n==0){
break;
}
for(int i=0;i<n;i++){
String s=sc.next();
map[i]=s.toCharArray();
}
boolean sign=false;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
if(map[i][j]=='@'){//先找到@这点，然后将其作为起点朝四个方向递归遍历
sign=true;//优化，因为只有一个@，所以只要进来，就不需要找循环找@了
DFS(i,j);
}
if(sign){//优化
break;
}
}
}
System.out.println(count+1);
}
}
private static void DFS(int a, int b) {
int x,y;
for(int i=0;i<4;i++){
x=a+dir[i][0];
y=b+dir[i][1];
if(x>=0&&y>=0&&x<n&&y<m&&map[x][y]=='.'){
count++;
map[x][y]='#';//直接把找到的'.' 变为#，这样就避免visit标记地图了
DFS(x,y);
}
}
}
}
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