题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1312

Red and Black

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 17773    Accepted Submission(s):
10826

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
题目大意： “.”代表黑地板，“#”代表红地板，“@”代表人的位置，人可以经过黑地板不能经过红地板，问在给定的图中人一共可以经过多少块地板
解题思路：深搜，在每个位置都搜索该位置的上下左右四个位置，将图中能够经过的点全部搜一遍
AC代码：

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <algorithm>

 #include <stack>

 #include <queue>

 using namespace std;

 int w,h;

 int a[][] = {{-,},{,},{,-},{,}};  //位置数组

 char str[][];

 int f[][];  //标记该位置是否走过

 int sum;

 void dfs(int x,int y)

 {

     f[x][y] = ;

     for (int i = ; i < ; i ++)

     {

         int x1=x + a[i][];

         int y1=y + a[i][];

         if (x1 >=  && x1 < h && y1 >=  && y1 < w && str[x1][y1]!='#' && f[x1][y1] == )

         {    //判断边界、是否是不能经过的红地板、该地板是否已经经过

             sum ++;

             dfs(x1,y1);

         }

     }

 }

 int main ()

 {

     int i,j,x,y;

     while (scanf("%d%d",&w,&h),w&&h)

     {

         for (i = ; i < h; i ++)

             scanf("%s",str[i]);

         memset(f,,sizeof(f));

         for (i = ; i < h; i ++)

         for (j = ; j < w; j ++)

             if (str[i][j] == '@') //找出人的位置

             {

                 x = i;

                 y = j;

                 break;

             }

         sum = ;

         dfs(x,y);

         printf("%d\n",sum);

     }

     return ;

 }

HDU 1312 Red and Black （dfs）的相关教程结束。