数据结构——HDU1312：Red and Black(DFS)

题目描述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.输入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.OutputFor 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

解题思路：
深搜的方法解决，题目意思就是从@开始找.并与@连通，碰到#等于碰到了墙，题目很简单，@可以向四个方向上、下、左、右走，所以 用四个坐标标记出来，然后，再一一遍历，递归调用寻找，用一个30*30的数组标识此点有没有走过，避免走重复

程序代码：

#include <cstdio>
#include <cstring>
using namespace std;
int n,m,cot;
char map[30][30];
int to[4][2] = {{1,0},{0,1},{-1,0},{0,-1}};

void dfs(int i,int j)
{
cot++;
map[i][j] = '#';
for(int k = 0; k<4; k++)
{
int x = i+to[k][0];
int y = j+to[k][1];
if(x<n && y<m && x>=0 && y>=0 && map[x][y] == '.')
dfs(x,y);
}
return;
}

int main()
{
int i,j,fi,fj;
while(~scanf("%d%d%*c",&m,&n)&&m&&n)
{
for(i = 0; i<n; i++)
{
for(j = 0; j<m; j++)
{
scanf("%c",&map[i][j]);
if(map[i][j] == '@')
{
fi = i;
fj = j;
}
}
getchar();
}
cot= 0;
dfs(fi,fj);
printf("%d\n",cot);
}

return 0;
}