思路：bfs裸题。
对这种迷宫问题的bfs，我们把坐标点用一个class来存储，并放入队列进行求解。
//一直接收不了输入，找了一个多小时的问题，居然是行和列搞反了ORZ
Red and Black
| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 48833 | Accepted: 26054 |
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)
The end of the input is indicated by a line consisting of two zeros.
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
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
class Point{
public int x;
public int y;
}
public class Main {
static int ans=0;//数量初始化为0
static int[][] dir={{0,1},{0,-1},{-1,0},{1,0}};//右左上下四个方向
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in=new Scanner(System.in);
int n=in.nextInt();//列数
int m=in.nextInt();//行数
Queue<Point> q=new LinkedList<Point>();//利用队列实现bfs
while(m!=0&&n!=0){//不为0时读入
char gra[][]=new char[m][n];//存迷宫
boolean vis[][]=new boolean[m][n];//标记是否访问
Point p=new Point();//表示坐标点位置
for(int i=0;i<m;i++){
String s=in.next();
for(int j=0;j<s.length();j++){
gra[i][j]=s.charAt(j);
if(gra[i][j]=='@'){ //找到@起始的位置
p.x=i;//x,y为搜索起点
p.y=j;
}
}
}
bfs(gra,vis,p,m,n,q);//调用
System.out.println(ans+1);//输出答案
ans=0;//恢复为0，为下一组做准备
q.clear();//清空队列
n=in.nextInt();//读入列
m=in.nextInt();//读入行
}
}
private static void bfs(char[][] gra, boolean vis[][],Point p, int m, int n,Queue q) {
// TODO Auto-generated method stub
vis[p.x][p.y]=true;//将起始点标记为已访问
q.add(p);//将起始点放入队列中
while(!q.isEmpty())//队列不为空时
{
Point qq=(Point) q.remove();//取出第一个元素
for(int i=0;i<4;i++)//枚举该点附近所有能到达的点
{
Point pp=new Point();
pp.x=qq.x+dir[i][0];
pp.y=qq.y+dir[i][1];
if((!(pp.x<0||pp.x>=m||pp.y<0||pp.y>=n))&&vis[pp.x][pp.y]==false&&gra[pp.x][pp.y]=='.')//如果点合法
{
vis[pp.x][pp.y]=true;//标记已访问
q.add(pp);//将该点加入队列中
ans++;//可达点数目+1
}
}
}}
}
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