HDU 1979 Red and Black

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
.#.........
.#.#######.
.#.#.....#.
.#.#.###.#.
.#.#..@#.#.
.#.#####.#.
.#.......#.
.#########.
...........
11 6
..#..#..#..
..#..#..#..
..#..#..###
..#..#..#@.
..#..#..#..
..#..#..#..
7 7
..#.#..
..#.#..
###.###
...@...
###.###
..#.#..
..#.#..
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：

 #include<stdio.h>
 char map[][];
 int dfs(int x,int y);
 int w,h;
 int main()
 {
     int i,j,sx,sy;
     while(scanf("%d%d",&w,&h),w+h != )
     {
         for(i=;i<=h;i++)
         {
             for(j=;j<=w;j++){
                 scanf(" %c",&map[i][j]);
                 if(map[i][j]=='@')
                 { sx=i;sy=j; }
             }
             getchar();
         }
         printf("%d\n",dfs(sx,sy));
     }
     return ;
 }
 int dfs(int x,int y)
 {
     if(x< || x>h || y< || y>w)
         return ;
     //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
     if(map[x][y]=='#')
         return ;
     else
     {
         map[x][y]='#';
         return +dfs(x-,y)+dfs(x+,y)+dfs(x,y-)+dfs(x,y+);
     }
 }

易错分析：

1、如果搜索进入不了注意边界的设置问题