Red and Black

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

题意：有一间房子，有红黑瓷砖，人只能在黑的上面活动，人在的位置就是@符号在的位置，求人能活动的瓷砖个数，人每次只能向前后左右移动；

解题思路：深搜，将走过的地板标记为1，具体过程看代码吧；

感悟：庆祝第一个独立写完的深搜代码！！！哦也！！！！

代码（G++
0ms）

#include

#include

#include

#include

#define maxn 100

using namespace std;

bool visit[maxn][maxn];

char mapn[maxn][maxn];

int direction[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

int w,h,s=1;

void dfs(int w,int h,int sx,int sy)

{

    int
x,y;

    for(int
i=0;i<4;i++)

    {

       
//printf("i=%d\n",i);

       
x=sx+direction[i][0];

       
y=sy+direction[i][1];//记录坐标

       
if(x>=1&&x<=h&&y>=1&&y<=w&&visit[x][y]==false&&mapn[x][y]=='.')

{

           
visit[x][y]=true;

           
s++;

           
dfs(w,h,x,y);

       
}

    }

}

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
x,y;

   
while(~scanf("%d%d\n",&w,&h)&&(w||h))

    {

       
//printf("w=%d h=%d\n",w,h);

       
s=1;

       
memset(visit,false,sizeof(visit));

       
for(int i=1;i<=h;i++)

       
{

           
for(int j=1;j<=w;j++)

           
{

               
scanf("%c",&mapn[i][j]);

              
// printf("mapn[i][j]=%c\n",mapn[i][j]);

               
//printf("i=%d\n",i);

               
if(mapn[i][j]=='@')

               
{

                   
x=i;

                   
y=j;

                   
visit[x][y]=true;

               
}//记录@符出现的位置

           
}

           
scanf("\n");

       
}

       
//for(int i=1;i<=h;i++)

       
//{

       
//    for(int
j=1;j<=w;j++)

       
//    {

       
//       
printf("%c",mapn[i][j]);

       
//    }

        
//   printf("\n");

       
//}

       
//printf("mapn[x][y]=%c\n",mapn[x][y]);

       
dfs(w,h,x,y);//开始深搜

       
printf("%d\n",s);

    }

    return
0;

}