HDU1312：Red and Black

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

 

 dfs搜索，懒得解释，自己看代码。

 

 

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n,m,ans;
char map[25][25];
int yi[4][2]={{1,0},{-1,0},{0,1},{0,-1}};
void dfs(int x,int y)
{
    int i,rx,ry;
    map[x][y]='#';
    ans++;
    for (i=0;i<4;i++)
    {
        rx=x+yi[i][0];
        ry=y+yi[i][1];
        if (rx>=1&&rx<=n&&ry>=1&&ry<=m&&map[rx][ry]=='.')
        {
            dfs(rx,ry);
        }
    }
}
int main()
{
    int i,j,x,y;
    while (~scanf("%d%d",&m,&n))
    {
        if (n==0&&m==0) break;
        ans=0;
        for (i=1;i<=n;i++)
        for (j=1;j<=m;j++)
        {
            scanf(" %c",&map[i][j]);
            if (map[i][j]=='@')
            {
                x=i;
                y=j;
            }
        }
        dfs(x,y);
        printf("%d
",ans);
    }
    return 0;
}