题目：
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 .#......... .#.#######. .#.#.....#. .#.#.###.#. .#.#[email protected]#.#. .#.#####.#. .#.......#. .#########. ........... 11 6 ..#..#..#.. ..#..#..#.. ..#..#..### ..#..#..#@. ..#..#..#.. ..#..#..#.. 7 7 ..#.#.. ..#.#.. ###.### [email protected] ###.### ..#.#.. ..#.#.. 0 0
Sample Output
45 59 6 13 题意描述：输入矩阵的大小W和H（均小于20）计算并输出从‘@‘位置最多能走多少块‘.‘解题思路：输入的时候找到‘@‘的位置，随后对其进行DFS搜索，下面的代码实现的搜索有点模拟广搜的意思。代码实现：
1 #include<stdio.h>
2 char map[30][30];
3 int dfs(int x,int y);
4 int w,h;
5 int main()
6 {
7 int i,j,sx,sy;
8 while(scanf("%d%d",&w,&h),w+h != 0)
9 {
10 for(i=1;i<=h;i++)
11 {
12 for(j=1;j<=w;j++){
13 scanf(" %c",&map[i][j]);
14 if(map[i][j]==‘@‘)
15 { sx=i;sy=j; }
16 }
17 getchar();
18 }
19 printf("%d\n",dfs(sx,sy));
20 }
21 return 0;
22 }
23 int dfs(int x,int y)
24 {
25 if(x<1 || x>h || y<1 || y>w)
26 return 0;
27 //如果进入不了dfs函数就是边界问题，注意行数和列数就是x和y的范围
28 if(map[x][y]==‘#‘)
29 return 0;
30 else
31 {
32 map[x][y]=‘#‘;
33 return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y-1)+dfs(x,y+1);
34 }
35 }
易错分析：
1、如果搜索进入不了注意边界的设置问题