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[30][30];
int dfs(int x,int y);
int w,h;
int main()
{
    int i,j,sx,sy;
    while(scanf("%d%d",&w,&h),w+h != 0)
    {
        for(i=1;i<=h;i++)
        {
            for(j=1;j<=w;j++){
                scanf(" %c",&map[i][j]);
                if(map[i][j]=='@')
                { sx=i;sy=j; }
            }
            getchar();
        }
        printf("%d
",dfs(sx,sy));
    }
    return 0;
}
int dfs(int x,int y)
{
    if(x<1 || x>h || y<1 || y>w)
        return 0;
    //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 0;
    else
    {
        map[x][y]='#';
        return 1+dfs(x-1,y)+dfs(x+1,y)+dfs(x,y-1)+dfs(x,y+1);
    }
}

易错分析：

1、如果搜索进入不了注意边界的设置问题