# HDU 1312 Red and Black

## 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


## AC 代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;
char mapp[25][25];
bool visit[25][25];
int s,w,h;;
void DFS(int x,int y)
{
if(x>=0&&x<h&&y>=0&&y<w&&mapp[x][y]!='#'&&!visit[x][y]) //如果在地图内并且该电是‘.’并且没有遍历过
{
s++;
visit[x][y]=true;                //标记已遍历
DFS(x+1,y);                      //四个方向
DFS(x,y+1);
DFS(x,y-1);
DFS(x-1,y);
}
}
int main()
{
while(cin>>w>>h&&(w||h))
{
int x,y;
s=0;
memset(visit,0,sizeof(visit));  //初始化visit数组
for(int i=0; i<h; i++)
{
cin>>mapp[i];
for(int j=0; j<w; j++)      //对每一行查找起点
if(mapp[i][j]=='@')x=i,y=j;
}
DFS(x,y);                       //深度优先遍历
cout<<s<<endl;
}
return 0;
}