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;
}